Electrostatics by Pramod Maheshwari

ELECTROSTATICS

EE R

JEE ADVANCED SYLLABUS

Coulomb's law

Electric field and potential

electrical potential energy of a system of point charges and of electrical dipoles in a uniform

Electric field lines

electrostatic field

CAREER POINT ___________________________________________________________ Electrostatics

SOLVED EXAMPLES The two forces are opposite and collinear. For equilibrium the forces must be equal, opposite and collinear. Hence 1 1 1 q . . = 4 0 rA 2 4 0 (2rA ) 2

1 in magnitude of either charge. 4 It can also be shown that for the equilibrium of B, the magnitude of q must be 1/4 of the magnitude of either charge. Stability. If q is slightly displaced towards A, FqA increases in magnitude while FqB

or q =

Sol.

The negative point charges of unit magnitude and a positive point charge q are placed along straight line. At what position and for what value of q will the system be in equilibrium ? Check whether it is stable, unstable or neutral equilibrium. The two negative charges A and B of unit magnitude are shown in fig. Let the positive charge q be at a distance rA from A and at

Ex.1

a distance rB from B. Force on q due to A A rA

–1 rB

Force on q due to A FqA =

1 q . 2 towards A 4 0 rA

Force on q due to B

bring the charge to its original position. So for perpendicular displacement, the equilibrium is stable. +q

1 q towards B. . 4 0 rB 2

EE R

FqB =

decreases in magnitude. Now the Net force on q is toward A so it will not return to its original position. So for axial displacement, the equilibrium is unstable. If q is displaced perpendicular to AB, the force FqA and FqB

These two forces acting on q are opposite and collinear. For the equilibrium of q. the two forces must also be equal i.e. | FqA | = | FqB | or

FqA

1 1 q q = . . 4 0 rA 2 4 0 rB 2

rA = rB

Hence

So for the equilibrium of q, it must be equidistant from A & B i.e. at the middle of AB Now for the equilibrium of the system, A and B must be in equilibrium. For the equilibrium of A 1 q Force on A by q = . towards q 4 0 rA 2

1 (1)(1) Force on A by B = . 4 0 ( rA  rB ) 2

1 1 . away from q 4 0 (2rA ) 2

Ex.2

Sol.

FqB

B –1

Charges of magnitude 100 micro coulomb each are located in vacuum at the corners A, B and C of an equilateral triangle measuring 4 metres on each side. If the charge at A and C are positive and the charge B negative, what is the magnitude and direction of the total force on the charge at C ? The situation is shown in fig. Let us consider the forces acting on C due to A and B. The force of repulsion on C due to A i.e., FCA in direction AC is given by

CAREER POINT ___________________________________________________________ Electrostatics

+q A a

–q

C 60º

FCA =

FCA

qq 1 . along AC 4 0 a2

1 qq = . along CB 4 0 a2

Substituting the given data, we get 1 1 1 × 40 × 10–6 × × 2 2 2

1 1  = 9 × 109 × 10–8 × 5 × 10–9     r 10  10 2 

1 5  10 6 100 – 10 = = 8 r 9 9  5  10

1 F  2  m r

A ball of mass 5 g and charge 10–7 C moves from point A whose potential is 500 V to a point B whose potential is zero. What is the velocity of the ball at the point A, if at the point B, it is 25 cm per second ? Let u be the velocity of the ball at point A. Workdone on the charge by the field W = q (VA – VB) = 10–7 (500 – 0) = 5 × 10–5 J This appears in the from of the increased K.E.

EE R

A particle of mass 40 mg and carrying a charge 5 × 10–9 C is moving directly towards fixed positive point charge of magnitude 10–8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest ? Is the acceleration constant during motion ? If the particle comes to rest momentarily at a distance r from the fixed charge, from 'conservation of energy' we have 1 1 Qq 1 Qq mu2 + = 4 0 a 4 0 r 2

Sol.



Ex.3

Ex.4

9  10 9  (100  10 6 ) 2

46a 2 42 = 5.625 Newton This force is parallel to AB.

acc. =

1 Qq 4 0 r 2

i.e., acceleration is not constant during motion.

FCA 2  FCB2  2FCA  FCB cos120 º q2

F =

Thus the two forces are equal in magnitude. The angle between them is 120º. The resultant force F is given by F=

As here so

The force of attraction on C due to B i.e., FCB in direction CB is given by FCB

r = 4.7 × 10–2 m

FCB

i.e., +q

1 100 190 = + 10 = m r 9 9

W =

1 1 mv2 – mu2 2 2

5 × 10–5 =

2  1 5  1  .    u 2  2 1000  4   

2 × 10–2 =

1 – u2 16

u2 =

1 1 17 – = 16 50 400

4.12 = 0.206 m/s 20 = 20.6 cm/sec.

u =

Ex.5

A pitch ball covered with tin foil having a mass of m kg hangs by a fine silk thread  metre long in an electric field E. When the ball is given an electric charge of q coulomb, it stands out d metre from the vertical line. Show that the electric field is given by E = mg d/q (  2  d 2 ) .

Electrostatics __________________________________________________________CAREER POINT

Sol.

As shown in fig. the following forces act on the ball when it is in equilibrium  

Tcos

Tsin

Sol.(i) forces acting on each bead are (a) mg downward (b) T, tension (c) electric force F (d) normal reactions. The directions of the forces are shown in the figure.



 Tsin N1

sA Tco 90º 

F mg

or But,

E = tan  =

c Mg

T sin  = F sin  + mg cos 30º (3)

N2 + F cos  = T cos  + mg cos 60º (4)

( 2  d 2 )

From (1) and (3) (T – F) cos  = mg sin 30º

d ( 2  d 2 )

and (T – F) sin = mg cos 30º Dividing, cot  = tan 30º = cot 60º    = 60º (ii) Substituting the value of  in (1) and noting that F =

and q2 are connected by a cord of length 

and can slide without friction on the wires. Considering the case when the beads are stationary, determine (i) the angle  = APQ, (ii) the tension in the cord, and (iii) the normal reactions on the beads. (iv) If the cords is now cut, what are the values of the charges for which the beads continue to remain stationary ? A q1  P 

0º os 3

Equating the forces along and perpendicular to AB and AC, for equilibrium of the beads T cos  = F cos  + mg sin 30º (1) and

A rigid insulated wire frame in the form of a right-angled triangle ABC, is set in a vertical plane as shown. Two beads of equal masses m and carrying charges q1

30º

F sin  + N1 = mg cos 30º + T sin  (2)

EE R



Ex.6

mg . tan  q

mg E = q

s T Fco in T s mg g m  n i s T mg 30º

(i) weight mg acting vertically downward (ii) tension T in the thread (iii) electric force q E horizontally to the right. From figure T sin  = qE and T cos = mg qE  tan  = mg

q2 Q 60º

T cos 60º = or

1 q1 q 2 4 0  2

1 q1 q 2 cos 60º + mg sin 60º 4 0  2

T =

1 q1 q 2 + mg 4 0  2

(iii) from (1)

From (2),

mg sin 30 º = mg cos 60 º N1 = mg cos 30º + (T – F) sin 60º



N1 = mg

T–F =

3 3 + mg . = 3 mg 2 2

From (4), N2 = (T – F) cos 60º ± mg cos 60º C

= mg ×

1 1 + mg × = mg 2 2

CAREER POINT ___________________________________________________________ Electrostatics

(iv) When the cord is cut, T = 0, then from (1) 0 = F cos 60º + mg sin 30º or F + mg = 0 This result is on the assumption that q1 and q2 are of the same sign, so it was taken

Sol.

The situation is shown in fig. Here the sum of kinetic energy and potential energy at B is the same as the sum of kinetic energy and potential energy at A i.e., qE

that there was a force of repulsion. Since mg is fixed in direction, its sign cannot be reversed but the sign of F can be reversed because if q1 and q2 are of opposite sign, F

q1q2 = 40 mg 2

REFERENCE LEVEL

E T1 mg O T2 qE

will change its sign from + to –. Let q1 and q2 be of the opposite sign, then –F + mg = 0 is the condition for equilibrium 1 q1 q 2 = mg or 4 0  2

(K.E.)B + (P.E.)B = (K.E.)A + (P.E.)A

..(1)

Gain in K.E. Ex.7

Sol.

EE R

The force experienced by the electron is F = Ee = 106 (1.6 × 10–19) = 1.6 × 10–13 newton The acceleration of the electron is given by

1.6  10 13 F = m 9.1  10 31 = 1.8 × 1017 m/sec2 The initial velocity = zero. Let t be the time taken by the electron in attaining a final speed of 0.1 c. Now v = u + at or v = at

a =



Ex.8

1 m (v22 – v12) ...(2) 2 P.E. = (P.E.)B – (P.E.)A

An electron (mass m charge e) is released from rest in a uniform electric field of 106 newton/coulomb. Compute its acceleration. Also find the time taken by the electron in attaining a speed of 0.1 c, where c is the velocity of light. (m = 9.1 × 10–31 kg, e = 1.6 × 10–19 coulomb and c = 3 ×108 metre/sec).

v 0.1  c 0.1  (3  10 8 ) t= = = a a 1.8  1017

= 1.7 × 10–10 sec.

A ball of mass m with a charge q can rotate in a vertical plane at the end of a string of length  in a uniform electrostatic field whose lines of force are directed upwards. What horizontal velocity must be imparted to the ball in the upper position so that the tension in the string in the lower position of the ball is 15 times the weight of the ball ?

= (K.E.)A – (K.E.)B = Loss in

= Loss in gravitational P.E. – gain in electric P.E. = mg . 2 – qE. 2 = (mg – qE) 2 ...(3) From eq. (1), (P.E.)B – (P.E.)A = (K.E.)A – (K.E.)B

Substituting the values, we get 1 (mg – qE) 2 = m (v22 – v12) ...(4) 2 Centripetal force at A = T2 + qE – mg 2

mv 2 ...(5)  From eq. (4) mv22 = 2 (mg – qE) 2 + mv12

 T2 + qE – mg =

From eq. (5) mv22 =  (T2 + qE – mg)  2(mg – qE) 2 – mv12 =  (T2 + qE – mg) m or 4mg – 4qE +   v12    = T2 + qE – mg

Putting

T2 = 15 mg, we have

m 4mg – 4qE +   v12 = 15 mg + qE – mg    m 2   v1 = 10mg + 5qE = 5 (2mg + qE)   

1  v1 =  .(2mg  qE) . 5 m 

1/ 2

Electrostatics _________________________________________________________CAREER POINT

x =01

V =

dVP = Y B

1m P

0.44 m



or 

r2 = x2 + y2

2 rdr = 2x dx dVP =

EE R

VP = =

q q 1 q   2  2  ... E =  2 4 0 1 2 4 

i.e.,

q 1 1 4  = q 4 0 [1  (1 / 4 )] 4 0  3 



VA =

Ex.10 On a thin rod of length  = 1 m, lying along

K 4 0



K [ 1.44 – 4 0

VB =

K [ 2 – 1] 4 0

= 0.4142

The situation is shown in fig. Consider a small element of length dx of the rod at a

K 4 0

= (9 × 109) × 10–9× 0.4142

10–9 cm–2. Find the workdone in displacing

Sol.

K 4 0

= 4.83 V

unit length  = K x, where K = constant =

0.44 m ) to (0, 1 m).

0.44 ]

= (9 × 109) × 10–9 × 0.5366

there is uniformly distributed charge per

(0,

K [ (  2  y 2 ) – y] 4 0

the x-axis with one end at the origin x = 0,

a charge q = 1000 C from a point

( 2  y2 )

= 0.5366

Ans.

K r dr K = dr 4 0 4  0 r

potential at point P. Thus

Ans.

1  1 1  1  4  16  64 ....  

(y = constant)

Integrating this expression, we get the

i.e.,

q = 4 0

From figure,

 1 1 1  1  2  4  8 ......  

1 q 1 = × (2q) 4 0 [1  (1 / 2)] 4 0

1 q q q     ... V =  4 0  1 2 4 

1 dx Kxdx = 4 0 r 4 0 r

interaction i.e., principle of superposition holds good -

q 4 0

y A

1 q 4 0 x

and electric interaction is a two body

dVP at point P due to this element is given

1 q E = 4 0 x 2

and

distance x from the origin. Then potential

Sol.

An infinite number of charges each equal to q are placed along the x-axis at x = 1, x = 2, x = 4, x = 8 ...... and so on. Find the potential and the electric field at the point x = 0 due to this set of charges. As potential V and intensity E due to a point charge at position x are respectively given byq q q q q

Ex.9

= 3.728 V Net workdone

= q (VB – VA) = 1000 × 10–6 [3.728 – 4.83] = – 1.1 × 10–3 Joule.

CAREER POINT __________________________________________________________Electrostatics

EXERCISE (Level-1) Q.1

Coulomb law & superposition

Q.5

Three charge +4q, Q and q are placed in a straight line of length  such that +q & +4q are on the ends of line & Q is at the mid point of line. What should be the value of Q in order to make the net force on q to be zero ? (A) – q (B) –2 q (C) –q/2 (D) 4q

Q.6

Four charges are arranged at the corners of a square ABCD, as shown. The force on a +ve charge kept at the centre of the square is C B –q +q

Four point charges are placed in a straight line with magnitude and separation as shown in the diagram. What should be the value of q0 such that + 10C charge is in equilibrium ?

+40µC

20cm 20cm

+10µC –10µC

(A) – 80 µC (C) + 80 µC Q.2

(B) + 40 µC (D) – 20 µC

Two small balls each having equal positive charge Q are suspended by two insulating strings at equal length L metre, from a hook fixed to a stand. The whole set-up is taken in a satellite into space where there is no gravity. Then the angle  between two strings and tension in each string is –

(A) 0,

kq 2 4L2

Q.7

(B) ,

Q.3

kq 2

CA Q.4

 kq 2 (D) , 2 2L2

Two point charges placed at a distance r in air exert a force F on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is – (A) r (B) r/8 (C) r/4 (D) r/2

+2q

Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be A –q B 3q

2L2

Two identical balls A and B having equal charges are placed at a fixed distance experience a force F. A similar uncharged ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is – (A) F/2 (B) F (C) 2F (D) 4F

–2q

(A) zero (B) along diagonal AC (C) along diagonal BD (D) perpendicular to the side AB

kq 2

EE R

Stand

40cm

Questions based on

–2q

(A) zero (C) along OC Questions based on

Q.8

D (B) along OF (D) none of these

Electric field A proton and an electron are placed in a same uniform electric field. (A) The electric forces acting on them will be equal (B) The magnitudes of the forces will be equal (C) Their accelerations will be equal (D) The magnitudes of acceleration will be equal

Electrostatics _________________________________________________________CAREER POINT

8r 3 g 3e

(B)

r 3 g 3e

(C)

6r 3 g 15e

(D)

4r 3 g 3e

Four charges q, 2q, – 4q and 2q are placed in order at the four corners of a square of side b. The net field at the centre of the square is – q (A) from + q to – 4q 2 0 b 2 (B) (C) (D)

Q.11

5q 2 0 b 2 10q 2 0 b 2

from + q to – 4q

20q 2 0 b 2

from – 4q to + q

Electrostatic potential and potential energy

–q Q (A) A, B, C, P and Q (B) A, B and C (C) A, P, C and Q (D) P, B and Q

–q

Q.12

Q.13

Q.15

(B) – 120 V (D) – 1200 V

A particle of mass 2 g and charge 1 µC is held at rest on a frictionless horizontal surface at a distance of 1 m form a fixed charge 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at distance of 10 m from the fixed charge is (A) 100 m/s (B) 90 m/s (C) 60 m/s (D) 45 m/s

Q.16

In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges are placed at (n – 1) corners. At the centre, the intensity is E and the potential is V. The ratio V/E has magnitude (A) r n (B) r (n – 1) (C) (n – 1)/r (D) r (n – 1)/n

Q.17

Figure shows equi-potential surface for a two charges system. At which of the labeled points will an electron have the highest potential energy ?

Figure represents a square carrying charges +q, +q, –q, – q at its four corners as shown. Then the potential will be zero at points P +q +q A

12 J of work has to be done against an existing electric field to take a charge of 0.01 C. from A to B. How much is the potential difference VB – VA? (A) 1200 V (C) 1250 V

from + q to – 4q

EE R

Questions based on

Q.14

(A)

between two points separated by a distance of 1 cm along field lines is (A) 10 volt (B) 90 volt (C) 1000 volt (D) 3000 volt

Q.10

A charged water drop of radius r is in equilibrium in an electric field. If charge on it is equal to charge on an electron, then intensity of electric field will be : (density of water = d, assume gravity)

Q.9

At a point situated at certain distance from a point charge the electric field is 500 V/m and the potential is 3000 V. What is the distance of the point from point charge ? (A) 6 m (B) 12 m (C) 36 m (D) 144 m When a charge of 3 coulomb is placed in a uniform electric field it experiences a force of 3000 Newton. The potential difference

-q

D C

(A) Point A (C) Point C

(B) Point B (D) Point D

CAREER POINT __________________________________________________________Electrostatics

Q.18

Relationship between Electric field & Potential Let V0 be the potential at the origin in an electric field E = Ex i + Ey j . The potential at the point (x,y) is (A) V0 – xEx – yEy

(A) 10 Vm–1 along PQ

(D ( x  y )

(B) 15 2 Vm–1 along PA  E2y

E 2x

D 0.2 m

P A

(B) V0 + xEx + yEy 2



0.2 m

Questions based on

–V0

Q.19

At any point (x, 0, 0) the electric potential  1000 1500 500  V is   2  3  volt , then x x   x

The equation of an equipotential line in an electric field is y = 2x, then the electric

electric field intensity at x = 1 m is -

field strength vector at (1, 2) may be -

(A) 5500 ĵ  k̂ V/m

(A) 4 î + 3 ĵ

(B) 4 î + 8 ĵ

(D) – 8 î + 4 ĵ





(B) 5500 îV/m 5500 (C) ĵ  k̂ V / m 2 5500 î  k̂ V / m (D) 2

 

Q.20

Q.22

(D) 5 Vm–1 along PA

 

Questions based on

Q.23

EE R

Figure below shows two equipotential lines in xy-plane for an electric field. The scales are marked. Electric field in the space between these equipotential lines are respectively – y t

ol 2v

The arc AB with the centre C and the infinitely long wire having linear charge density  are lying in the same plane. The minimum amount of work to be done to move a point charge q0 from point A to B through a circular path AB of radius a is equal to –

olt 4v

y (in cm)

Electric field intensity & potential due to different charge distribution

+

+ + + + + + + + + +

x 8 x (in cm)

(A) + 100 î – 200 ĵ V/m

(B) – 100 î + 200 ĵ V/m

A, B, C, D, P and Q are points in a uniform electric field. The potentials at these points are V (A) = 2 volt. V (P) = V (B) = V (D) = 5 volt. V (C) = 8 volt. The electric field at P is -

–

(D) – 200 î – 100 ĵ V/m

Q.21

B  a

(A)

q 20 2 loge   2 0 3

(B)

(C)

q0  2 loge   2 0 3

(D) q0  / 2 0

q0  3 loge   2 0 2

Electrostatics _________________________________________________________CAREER POINT

The linear charge density on upper half of a segment of ring is  and at lower half it is –. The direction of electric field at centre O of ring is : A + + + + B D O – – – – C

Q.25

The electric potential due to an infinite sheet of positive charge density at a point located at a perpendicular distance Z from the sheet is: (Assume V0 to be the potential at the surface of sheet) Z (B) V0 –  0

(A) V0 (C) V0 +

Z 2 0

(D) V0 –

Z 2 0

120º

(A)

(B)

(C)

(D)

Point P lies on the axis of a dipole. If the dipole is rotated by 90º anti clock wise, the  electric field vector E at P will rotate by (A) 90º clock wise (B) 180º clock wise (C) 90º anti clock wise (D) none of these

Q.29

A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. A and B are far away from the dipole and at equal distance from it.   The field at A and B are E A and E B .   (A) E A = E B   (B) E A = 2 E B   (C) E A = – 2 E B    1  (D) | E B | = | E A | and E B is  to E A 2

Q.30

Two short electric dipoles are placed as shown. The energy of electric interaction between these dipoles will be -

Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is -

EE R

Q.26

Q.28

(B) along OB (D) along OD

–

(A) along OA (C) along OC

Q.24

+3Q

120º

–2Q

Questions based on

Q.27

(A)

Q 4 0 R

(B)

Q 2 0 R

(C)

2Q  0 R

(D)

Q  0 R



r  



Electric dipole Figure shows the electric field lines around and electric dipole. Which of the arrows best represents the electric field at point P?

(A) (C)

2kP1 P2 cos  3

r 2kP1 P2 sin  r

(B) (D)

2kP1 P2 cos  r3

4kP1P2 cos  r3

CAREER POINT __________________________________________________________Electrostatics

EXERCISE (Level-2) Q.5

There are two point charges q1 and q2 lying on a circle of unit radius. Electric field intensity at the center of circle due to these  charges is E . Find the position vector of the center with respect to q2 if the position  vector of the center respect to q1 is r1 .      E  kq 1 r1  E  kq1 r1   (A) (B) –  kq 2  kq 2       kq 1 r1  E E  kq 1 r1 (C) (D) kq 2 kq 2

Q.2

(A) 2 (C) 3 Q.6



position

vector

r  2î  3 ĵ  6k̂

electric field will be -

then

(A)

(2î  3 ĵ  6k̂ )K 243

(B)

(2î  3 ĵ  6k̂ )K 343

(C)

(3î  2 ĵ  6k̂ )K 243

(D)

(6î  2 ĵ  3k̂ )K 343

Q.7

In given figure, a point charge +q is placed at the origin O. Work done in taking another point charge –Q from the point A [co-ordinates (0, a)] to another point B [coordinates (a,0)] along the straight path AB is-

EE R

Q.3

A square of side b centered at the origin with side parallel to axes of x and y has surface charge density (x, y) = 0xy within its boundaries. Total charge on the square is (A) 0 (B) 0b2 (C) 20b2 (D) 40b2

Electric potential in an electric field is given as V = K/r, (K being constant). At

(B) 1 (D) none of these

Q.1

Five positive equal charges are placed at vertices of a regular hexagon and net electric field at the center is E1. A negative charge having equal magnitude is placed sixth vertex and then net electric field is E E2. Find 2 . E1

Single correct answer type questions

Y A

Two parallel plates of infinite dimensions are uniformly charged. The surface charge density on one is A and on the other is B, field intensity at point C will beD

–

(A) Proportional to( A – B ) (B) Proportional to( A + B ) (C) Zero (D) 2A

Q.4

–

– 

Q.8

(A) Zero

  qQ 1   2a (B)  2   4  0 a 

 qQ 1  a  (C)  2   4  0 a  2

 qQ 1   2a (D)  2   4  0 a 

–

B–



An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its A charged particle of mass (m) is kept in dipole moment is along the direction of the equilibrium in the electric field between field, the force on it and its potential the plates of Millikan oil drop experiment. energy are respectivelyIf the direction of the electric field between (A) 2q.E and minimum the plate is reversed, then acceleration of (B) q.E and p.E the charged particle will be (C) Zero and minimum (A) Zero (B) g /2 (C) g (D) 2g (D) q.E and maximum _________________________________________________________ Electrostatics CAREER POINT

The variation of electric potential with distance from a fixed point is shown in figure. What is the value of electric field at x = 2m – Y 6 V(Volt)

4 2 0

Q.11

EE R

(A) Zero

1 2 3 4 X x(m) (B) 6/2 (C) 6/1 (D) 6/3

1 q q0 4  0 3 R

(D)

(B) 6 î  8 ĵ

(D) Zero

Q.15

Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third uncharged spherical conductor having same radius as that of B is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is(A) F/4 (B) 3F/4 (C) F/8 (D) 3F/8

Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential(A) Continuously increases (B) Continuously decreases (C) Increases then decreases (D) Decreases then increases

CA Qqm  0 r

(D)

Q.16

A small ball of mass m and charge + q tied with a string of length , is rotating in a vertical circle under gravity and a uniform horizontal electric field E as shown. The tension in the string will be minimum for -

1 q q0 4  0 3 R

A charged particle of charge Q is held fixed and another charged particle of mass m and charge q (of the same sign) is released from a distance r. The impulse of the force exerted by the external agent on the fixed charge by the time distance between Q and q becomes 2r is Qq Qqm (B) (A) 4  0 mr 4  0 r (C)

(A)  6 î  8 ĵ

3 R from its centre. How much

kinetic energy will be acquired by the test charge when it reaches the centre of the ring – 1 qq0 1 qq0 (A) (B) 4  0 R 4  0 2 R (C)

Electric potential is given by : V = 6x – 8xy2 – 8y + 6yz – 4z2 Electric field at the origin is -

Q.14

A ring of radius R carries a charge +q. A test charge –q0 is released on its axis at a distance

Q.12

Q.13

Q.10

An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20cm from this origin such that OP makes  an angle with the x-axis. If the electric 3 field at P makes an angle  with the x-axis, the value of  would be –  3    (A) (B)  tan 1   2  3 3    3 2  (C) (D) tan 1   2  3  

Q.9

Qqm 2 0 r





E m

 qE  (A)  = tan–1    mg  (B)  =   qE  (C)  =  – tan–1    mg   qE  (D)  =  + tan–1    mg 

CAREER POINT __________________________________________________________Electrostatics

The variation of electric field between the two charges q1 and q2 along the line joining the charges is plotted against distance from q1 (taking rightward direction of electric field a positive) as shown in the figure. Then the correct statement is – E

(A)

EE R

Q.20

A wheel having mass m has charges +q and –q on diametrically opposite points. It means in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E = +q E –q 

(A)

mg q

(B)

mg 2q

(C)

mg tan  2q

(D) none

q

1 2 (C) –2q

Q.22

(B)

2q

2 2 (D) +q

Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in –

(A)

(B)

(C)

(D)

Q.23

A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = + 1 cm and C be the point on the y axis at y = + 1 cm. Then the potentials at the points A, B and C satisfy (A) VA < VB (B) VA > VB (C) VA < VC (D) VA > VC

Q.24

Two equal point charges are fixed at x = – a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to – (A) x (B) x2 (C) x3 (D) 1 / x

When two charges are placed at distance a apart. Find the magnitude of third charge which is placed at mid point the line joining the charge. So that system is in equilibrium Q Q Q (A)  (B)  (C)  (D) – Q 4 2 3

(A) q1 and q2 are positive and q1 < q2 (B) q1 and q2 are positive and q1 > q2 (C) q1 is positive and q2 is negative and q1 < q2 (D) q1 and q2 are negative and q1 < q2 Q.19

Three charges Q, +q and +q are placed at the vertices of a right angled isosceles triangle as shown in figure. The net electrostatics energy of the configuration is zero if Q is equal to – Q

Q.18

Q.21

Four charges equal to – Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is – Q Q (1 + 2 2 ) (B) (1 + 2 2 ) (A) – 4 4 Q Q (C) – (1 + 2 2 ) (D) (1 + 2 2 ) 2 2

Q.17

Electrostatics _________________________________________________________CAREER POINT

Six charges of equal magnitude each of q are placed at six corners of a regular hexagon. Find arrangement the charges in order PQRSTU which produce double electric field at centre O as compared to electric filed produce by single charge +q placed at R -

Q.29 S

(A) +++- - (C) - + + - + -

(B) + - + - + (D) - + + + - -

+ +

Charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged nonconducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable. (A) For all values of H R (B) Only if H > 2 R (C) Only if H < 2 R (D) Only if H = 2

Find the electric force on 2µC charge placed at the common centre of two equilateral triangles each of side 10 cm. as shown in figure. Electric charge on A, B, C, D, E and F points are + 2µC, +2µC, +2µC, –2µC, –2µC and –2µC respectively F

Q.27

EE R

Five balls numbered 1 to 5 are suspended using separate threads, Pairs (1, 2), (2, 4), (4, 1) shows electrostatics attraction, while pairs (2, 3) and (4, 5) show repulsion therefore ball 1 must be (A) Positively charged (B) Negative charged (C) Neutral (D) None

(A) Will increase (B) Will decrease (C) Will not change (D) Will first increase then decrease

Q.26

+ +

Q.25

(A) 64.8 N (C) Zero

Q.28

(B) 21.6 N (D) 43.2 N

If a positively charged pendulum is oscillating in a uniform electric field as shown in fig. Its time period as compared to that when it was uncharged.

Q.30

10C charge is uniformly distributed over a thin ring of radius 1m. A particle (mass = 0.9 gm, charge –1C) is placed at the centre of ring. It is displaced along the axis of ring by small displacement, then time period of SHM of particle (A) 0.6 sec (B) 0.2 sec. (C) 0.3 sec. (D) 0.4 sec.

Q.31

Four charges q1, q2, q3 and q4 are placed at the positions as shown in the figure, given q1 + q2 + q3 + q4 = 0. The electric field on z-axis y q (0,a) 2 q3

(–a,0)

(a,0)

q4 (0,–a)

(A) is always along +ve z-axis (B) is always along –ve z-axis (C) is always zero (D) may be perpendicular to z-axis or zero depending on charges

CAREER POINT __________________________________________________________Electrostatics

Figures shows four situations in which charged particles are at equal distances from the origin. If E1, E2, E3 and E4 be the magnitude of the net electric fields at the origin in four situations (i), (ii), (iii) and (iv) respectively, then y y –5q –5q

Q.35

Two point charges +8q and –2q are located at x = 0 and x = L respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero – L (A) 8L (B) 4L (C) 2L (D) 4

x (ii)

Q.36

Three charges –q1, +q2 and –q3 are placed as shown in the figure. The x-component of the force on –q1 is proportional to -

–3q

(iii)

–q y

–q x(iv) q

–2q

(i)

–4q

Q.32

–q3

(A) E1 = E2 = E3 = E4 (B) E1 = E2 > E3 > E4 (C) E1 < E2 < E3 = E4 (D) E1 > E2 = E3 < E4 Q.33

EE R

non conducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d << R). The electric field at the centre of the ring will now be (A) directed towards the gap, inversely proportional to R3 (B) directed towards the gap, inversely proportional to R2 (C) directed away from the gap, inversely proportional to R3 (D) directed away from the gap, inversely proportional to R2

Q.34

The figure shows three non conducting rods one circular and two straight. Each has a uniform charge of magnitude Q along its top half and another along its bottom half. Which of them correctly represents the direction of field at point P ?

(I) (A) I

(C)

Q.37

–Q (III)

–Q (II)

(B) II

q2 2

b q2 b2

Q.38

–

q3 2

a q3 a2

(D) I and II

+q2

–q1

sin 

(B)

cos  b a2 q q (D) 22 + 32 cos  b a 2

–

From a point if we move in a direction making an angle  measured from + ve x-axis, the potential gradient is given as dv = 2 cos . Find the direction and dr magnitude of electric field at that point (A) 2 î

(B) –2 î

Point charges q, – q, 2Q and Q are placed in order at the corners A, B, C, D, of a square of side 2b. If the field at the q midpoint of CD is zero, then is Q (A) 1

Q.39

–Q

(A)

(B) 2



(C)

2 2 5 5 (D) 5 2

A point charge 50 C is located in the XY plane at the point of position vector  r0 = 2 î + 3 ĵ . what is the electric field at the  point of position vector r = 8 î – 5 ĵ . (A) 1200 V/m (C) 900 V/m

(B) 0.04 V/m (D) 4500 V/m

Electrostatics _________________________________________________________CAREER POINT

(A) 0 V Q.44

x (3,0) (cm)

(B) 4V

In space of horizontal electric field (E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position –

E = mg

q, m

Q.45

(A)

g 

(B)

2g 

(C)

3g 

(D)

5g 

A charge Q is placed at the centre of a circle of radius R. The work done in moving a charge q from A to B so as to complete a semicircle is –

–q

R A

q, m

(D) none of these

Six charges of magnitude + q and – q are fixed at the corners of a regular hexagon of edge length a as shown in the figure. The electrostatic interaction energy of the charged particles is : +q –q

–q

q 2  3 15  q2  –  (B)    0 a  8 4   0 a  q 2  3 15  q2  (C) –  (D)    0 a  4 2   0 a 

(A)

45º

(B) zero

Q.42

EE R

(C)

(cm) y (0,2) A

–2q

9 3kq 8L2 9 kq

given by–

45º

(A)

A uniform electric field of 400 V/m is directed at 45º above the x-axis as shown in figure. The potential difference VA – VB is

An equilateral triangle wire frame of side L having 3 point charges at its vertices is kept in x-y plane as shown. Component of electric field due to the configuration in z direction at (0, 0, L) is [origin is centroid of triangle] -

Q.43

Q.41

A point charge q is placed at origin. Let    E A , E B and E C be the electric field at three points A (1, 2, 3), B (1, 1, –1) and C (2, 2, 2) due to charge q. Then   (i) E A  E B   (ii) | E B | = 4 | E C | select the correct alternative (A) only (i) is correct (B) only (ii) is correct (C) both (i) and (ii) are correct (D) both (i) and (ii) are wrong

Q.40

3 9 –  2 4  3 15  –  2 8 

(A) Zero (C)

Qq 2 0 R

(B) (D)

Qq 4 0 R Qq 4 0 R 2

CAREER POINT __________________________________________________________Electrostatics

Q.46

Point charge (q) moves from point (P) to point (S) along the path PQRS as shown in fig. in a uniform electric field E, pointing parallel to the positive direction of the xaxis. The co-ordinates of the points P,Q,R and S are (a, b, 0), (2a, 0, 0), (a, -b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression –

Q.49

The diagram shows a small bead of mass m carrying charge q. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge +Q has also been fixed as shown. The potential at the point P due to +Q is V. The velocity with which the bead should projected from the point P so that it can complete a circle should be greater than 

(A) (C)

Q.50

(D) qE [(2a ) 2  ( b ) 2 ] Q.47

EE R

In a certain charge distribution, all points having zero potential can be joined by a circle S. Points inside S have positive potential, and points outside S have negative potential. A positive charge, Which is free to move, is placed inside S. (A) It will remain in equilibrium (B) It can move inside S but it cannot cross S (C) It must cross S at some time (D) It may move, but will ultimately return to its starting point

Q.48

The potential field depends on x and y coordinates as V = (x2 – y2). Corresponding electric field lines in x-y plane as shown in Fig y y (A)

(B)

Bead

6qV m

(B)

3qV m

(D) None of these

(A) q E a (B) –q E a (C) q E b



+Q a

qV m

The diagram shows three infinitely long uniform line charges placed on the X, Y and Z axis. The work done in moving a unit positive charge from (1, 1, 1) to (0, 1, 1) is equal toY 3

X 2

 Z

(A) (ln 2) /2  (C) (3ln 2) / 2 Q.51

(B)ln (D) None of these

4 charges are placed each at a distance ‘a’ from origin. The dipole moment of configuration isy 3q 0, 0

–2q

x –2q

y (C)

(D)

(A) 2qaĵ

(B) 3qaĵ

(D) None of these

Electrostatics _________________________________________________________CAREER POINT

(C)

Q.53

0 V 

(D)

Q.55

2 0 V

The magnitude of electric field intensity at point B (2, 0, 0) due to a dipole of dipole  moment, P = î + 3 ĵ kept at origin is

(C)

7k 8

(B)

13k 4

(D)

7k 4

(B)

 ML 2 2qE

ML 6qE

(D) 4

ML 2qE

13k 8

ML 2qE

(A) 2

(assume that the point B is at large 1 )distance from the dipole and k = 4 0 (A)

A point particle of mass (M) is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of mass (2M) is attached to the other end of the rod. The two particles carry charges +q and -q respectively. This arrangement is held in a region of uniform electric field (E). Such that the rod makes a small angle ( 5º) with the field direction the minimum time needed for the rod to become parallel to the field after it is set free –

An infinite layer of charge has a surface charge density of ''C/m2. The separation between two equi-potential surfaces, whose potential differs by V volt is 2 0 V V (B) (A) 0 

Q.52

Q.56

A short dipole is placed at origin of coordinate system as shown in figure, find the electric field at point P (0, y). y

Find the angular position , where the electric field due to an electric dipole is perpendicular to the dipole moment –

EE R

Q.54

P 0, y P

45º

E 





(B) tan–1 2

(D) tan–1 7

(C)

KP y

(– î – 2 ĵ)

KP 2y 3

( î  2 ĵ)

(B)

KP 2y 3

( î  2 ĵ)

(D) None of these

(A) tan–1 3

(A)

CAREER POINT __________________________________________________________Electrostatics

EXERCISE (Level-3) Part-A : Multiple correct answer type questions

Equipotential surfaces (A) are closer in regions of large electric fields compared to regions of lower electric fields. (B) will be more crowded near sharp edges of a conductor. (C) will be more crowded near regions of large charge densities. (D) will always be equally spaced.

Q.4

An electron is placed just in the middle between two long fixed line charges of charge density +  each. The wires are in the xy plane (Do not consider gravity) + +

A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring. Then -

q R

(A) If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre. (B) If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring. (C) If q < 0, it will perform SHM for small displacement along the axis. (D) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0.

Q.1

Q.3

Four identical particles each having mass m and charge q are placed at the vertices of a square of side . All the particles are free to move without any friction and released simultaneously from rest. Then (A) At all instants, the particles remains at vertices of square whose edge length is changing. (B) The configuration is changing (not remaining square) as the time passes (C) The speed of the particles when one of  the particles get displaced by is 2 q2  1  2     8 0 m  2

(D) Speed of the particles can not be found

(A) The equilibrium of the electron unstable along x-direction (B) The equilibrium of the electron neutral along y-direction. (C) The equilibrium of the electron stable along z-direction (D) The equilibrium of the electron stable along y-direction

EE R

Q.2

Q.5

will be will be will be will be

Two positively charged particles X and Y are initially far away from each other and at rest. X begins to move towards Y with some initial velocity. The total momentum and energy of the system are p and E (A) If Y is fixed, both p and E are conserved. (B) If Y is fixed, E is conserved, but not p. (C) If both are free to move, p is conserved but not E. (D) If both are free, E is conserved, but not p.

Electrostatics _________________________________________________________CAREER POINT

Four short dipoles each of dipole moments are placed at the vertices of a square of side a. The direction of the dipole moments are shown in the figure. Find the electric field (E) and potential (V) at the centre ‘O’ of the square.

(A) periodic, for all value of z0 satisfying 0 < Z0 <  (B) simple harmonic, for all value of Z0 satisfying 0 < Z0  R (C) approximately provided Z0 << R

P P

(D) such that P crosses O and continues to move along the negative z-axis towards

Z = – 

Q.7

2 2P  0 a

2P  0 a

(B) V =

(D) V =

Q.10

 0 a 2 2 2P  0 a 2

Three charged particles are in equilibrium under their electrostatic forces only – (A) The particles must be collinear

(B) All the charges cannot have the same

+q A

magnitude

EE R

(D) The equilibrium is unstable

Two identical charges +Q are kept at fixed distance apart. A small particle P with charge q is placed midway between them. If P is given a small displacement , it will undergo simple harmonic motion if – (A) q is positive and  is along the line joining the charges (B) q is positive and  is perpendicular to the line joining the charges (C) q is negative and  is perpendicular to the line joining the charges (D) q is negative and  is along the line joining the charges

CA Q.9

A positively charged thin ring of radius R is fixed in the xy plane, with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0 Z0), where Z0 > 0. Then the motion of P is –

C +q

–q B

1 q2 along OE 4 0 h 2

(D) the resultant force on –q at O is

Q.8

(A) the forces on (–q) at O due to charges +q at A and D are balanced (B) the forces on (–q) due to charges at D and E are balanced (C) the resultant force on –q at O is

Five point charges, each of charge +q coulomb are placed on five vertices of a regular hexagon of side h as shown in figure. Then select the correct options – +q +q D E

(A) E =

harmonic

simple

Q.6

1 q2 along OC 4 0 h 2

Q.11

Four identical charges are placed at the points (1, 0, 0), (0, 1, 0), (–1, 0, 0) and (0, – 1, 0)– (A) The potential at the origin is zero (B) The field at the origin is zero (C) The potential at all points on the z-axis, other than the origin, is zero (D) The field at all points on the z-axis other than the origin, acts along the zaxis

CAREER POINT __________________________________________________________Electrostatics

Four charges, all of the same magnitude, are placed at the four corners of a square. At the centre of the square, the potential is V and the field is E. By suitable choices of the signs of the four charges, which of the following can be obtained – (A) V = 0, E = 0 (B) V = 0, E  0 (C) V  0, E = 0 (D) V  0, E  0

Q.13

Q.16

A particle A of mass m and charge Q moves directly towards a fixed particle B, which has charge Q. The speed of A is v when it is far away from B. The minimum separation between the particle is proportional to – (A) Q2 (B) 1/v2 (C) 1/ v (D) 1/m

Fig. I

10V

EE R

(0.6 î – 0.4 ĵ – 0.9k̂ ) N m.

(B) The potential energy of the dipole is – 0.6 J. (C) The potential energy of the dipole is 0.6 J. (D) If the dipole is rotated in the electric field, the maximum potential energy of the dipole 1.3 J.

50V

20V

40V

50V

Fig. III (A) The work done in figure (i) is the greatest. (B) The work done in figure (ii) is least. (C) The work done is the same in figure (i), figure (ii) and figure (iii). (D) The work done in figure (iii) is greater than figure (ii) but equal to that in figure (i).

(A) Maximum stretch in the spring is

 An electric dipole moment p = (2.0î  3.0 ĵ) C.m, is placed in a uniform electric field  E = (3.0î  2.0k̂ ) × 105 N C–1.   (A) The torque that E exerts on p is

10V 20V

30V

q,m

2qE k (B) In equilibrium position, stretch in the qE spring is k qE (C) Amplitude of oscillation of block is k 2qE (D) Amplitude of oscillation of block is k

Fig. II (ii)

10V 20V 30V 40V 50V

A block having mass m and charge q is connected by spring of force constant k. The block lies on a frictionless horizontal track and a uniform electric field E acts on system as shown. The block is released from rest when spring is unstretched (at x = 0). Then select the correct options -

Q.15

Q.14

Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B. 30V 40V

Q.12

Q.17

Variation of electrostatic potential along xdirection is shown in the graph. The correct statement/s about electric field is V

C x

(A) x component at point B is maximum (B) x component at point A is towards positive x-axis. (C) x component at point C is along negative x-axis (D) x component at point C is along positive x-axis

Electrostatics _________________________________________________________CAREER POINT

Q.22

A particle of charge 1C & mass 1 gm moving with a velocity of 4 m/s is subjected to a uniform electric field of magnitude 300 V/m for 10 sec. Then it's final speed cannot be : (A) 0.5 m/s (B) 4 m/s (C) 3 m/s (D) 6 m/s

Q.23

The figure shows a nonconducting ring which has positive and negative charge non uniformly distributed on it such that the total charge is zero. Which of the following statements is true ?

Two point charged Q and – Q/4 are separated by a distance x. Then. Q

Q.19

Select the correct statement/s : (Only force on a particle is due to electric field) (A) A charged particle always moves along the electric line of force. (B) A charged particle may move along the line of force (C) A charged particle never moves along the line of force (D) A charged particle moves along the line of force only if released from rest.

–Q/4

Q.18

+ – + – – + – + – + – + – O –– + + – – + – – + + –

Q.20

EE R

(A) Potential is zero at a point on the axis which is x/3 on the right side of the charge –Q/4 (B) Potential is zero at a point on the axis which is x/5 on the left side of the charge –Q/4 (C) electric field is zero at a point on the axis which is x on the right side of the charge –Q/4 (D) there exist two points on the axis where electric field is zero.

R CA Q.21

(A) The potential at all the points on the axis will be zero. (B) The electric field at all the points on the axis will be zero. (C) The direction of electric field at all points on the axis will be along the axis. (D) If the ring is placed inside a uniform external electric field then net torque and force acting on the ring would be zero.

Two fixed charge 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m. +4Q –Q A 3m B (A) The point P where the resultant field due to both is zero is on AB outside AB. (B) The point P where the resultant field due to both is zero is on AB inside AB. (C) If a positive charges is placed at P and displaced slightly along AB it will execute oscillations. (D) If a negative charge is placed at P and displaced slightly along AB it will execute oscillations. An electric charges 10–8C is placed at the point (4m, 7m, 2m). At the point (1m, 3m, 2m), the electric (A) potential will be 18 V (B) field has no Y-component (C) field will be along Z-axis (D) potential will be 1.8 V

axis

Part-B : Column Matching type Questions Q.24

In figure, charges, each + q, are fixed at L and M. O is the mid-point of distance LM. X and Y axes are as shown. Consider the situations given in column-I and match them with the information in column-II. Y

+q L

+q O

CAREER POINT __________________________________________________________Electrostatics

(C)

Q.28

Suppose particle Z has a charge of +2C, and it begins and ends at rest. If E is 5 N/C, how much work is done on particle Z – (A) 10 J (B) 20 J (C) 40 J (D) 80 J

Q.29

Suppose that the field strength E is 10 N/C and particle Y has a charge of –10 C. When particle Y is released from rest, it follows the path as shown and accelerates to a velocity of 10 m/s. What is the mass of particle Y– (A) 1 kg (B) 2 kg (C) 3 kg (D) 4 kg

(D)

If the particles are positively charged, which particles increased their electric potential energy(A) X and Z (B) Y and Z (C) W, X, Y and Z (D) Since the electric field is constant, none of the particles increased their electric potential energy

(B)

Q.27

(A)

Column-I Column-II Let us place a charge + q at (P) Force on the O, displace it slightly along charge is zero x-axis and release. Assume that it is allowed to move only along X-axis. At position O, Place a charge – q at O. (Q) Potential energy Displace it slightly along X-axis of the system is and release. Assume that it is maximum allowed to move only along X-axis. At position O, Place a charge + q at O. (R) Potential energy Displace it slightly along Y-axis of the system is and release. Assume that it is minimum allowed to move only along Y-axis. At position O, Place a charge – q at O. (S) The charge is Displace it slightly along Y-axis in equilibrium and release. Assume that it is allowed to move only along Y-axis. At position O,

Part-C : Passage based objective questions

EE R

Passage # 1 (Q. 25 to 29) In the diagram (given below), the broken lines represent the paths followed by particles W, X, Y and Z respectively through the constant field E. The numbers below the field represent meters. E

Passage # 2 (Q.30 to 35) A very large, charged plate floats in deep space. Due to the charge on the plate, a constant electric field E exists everywhere above the plate. An object with mass m and charge q is shot upward from the plate with a velocity v and at an angle . It follows the path shown reaching a height h and a range R. Assume the effects of gravity to be negligible .



Q.30

Which of the following must be true concerning the object – (A) q must be positive (B) q must be negative (C) m must be large (D) m must be small

Q.31

Which of the following gives the vertical velocity of the object in terms of h just before colliding with the plate at the end of its flight–

If the particles begin and end at rest, and all are positively charged, the same amount of work done on which particles – (A) W and Y (B) W, Y and Z (C) Y and Z (D) W, X, Y and Z

Q.26

If all particles started from rest and all are positively charged, which particles must have been acted upon by a force other than that produced by the electric field – (A) W and Y (B) X and Z (C) X, Y and Z (D) W, X, Y and Z

Q.25

(A) 2gh (B) 2Eqh (C)

2mh 2qhE (D) Eq m

Electrostatics _________________________________________________________CAREER POINT

Which of the following is true concerning all objects that follow the path shown when propelled with a velocity v at an angle  – (A) They must have the same mass (B) They must have the same charge (C) They must have the same mass and the same charge (D) Their mass to charge ratios must be the same.

Q.34

Which of the following will result in an increase in R – (A) increasing both q and m by a factor of 2 (B) decreasing both q and m by a factor of 2 (C) increasing q by a factor of 2 while decreasing m by a factor of 2 (D) decreasing q by a factor of 2 while increasing m by a factor of 2

Which of the following is true concerning the flight of the projectile shown – (A) Increasing the mass m decreases the maximum height h (B) Increasing the charge q increases the maximum height h (C) Increasing the mass m decreases the downward acceleration (D) Increasing the charge q decreases the downward acceleration

EE R

Q.35

15V 10V

Path-1 B A Path-2 5V

Student 1 – Pushing the charge along path 1 requires less work, because work is proportional to distance. Since path 1 is shorter, the work is less. Student 2 – But work also depends on force. Along path 1, you’re pushing the charge directly against the electric field. This requires a large force. By contrast, along path 2, you’re not fighting the field head on. Since you can push the particle with a much smaller force, the total work is less along path 2, even though more distance is covered.

Suppose E is 10 N/C, m is 1kg, q is – 1 C, v is 100 m/s and  is 30º. What is h – (A) 25 m (B) 45 m (C) 80 m (D)125 m

object

Q.33

Q.32

Passage # 3 (Q.36 to 39) An asymmetric object creates an electric field corresponding to the equipotential lines in figure 1. (These lines are a crosssection of the three dimensional equipotential surface). Each line differs in potential from its nearest neighbor by 5 volts. Electric field lines point perpendicular to equipotential lines – A positive point charge is going to be pushed from point A to B. Two students argue about which requires less work: Pushing the charge along path 1 or pushing it along path 2.

Q.36

Which student, if either, is correct about the work required to push the point charge along path 1vs.path2– (A) Student 1 (B) Student 2 (C) Neither student 1 nor student 2 (D) We cannot determine who is correct without more information

Q.37

What additional information is needed to calculate the work done while pushing the point charge at constant speed from A to B along path 1 – (i) The charge of the point charge (ii) The charge of the asymmetric blob (iii) The electric field along that path (A) (i) only (B) (i) and (ii) (C) (i) and (iii) (D) (i), (ii) and (iii)

Q.38

The point charge has charge 0.01 coulombs. How much work is needed to push it from A to C along the equipotential line containing those two points – (A) –0.2 J (B) 0 J (C) 0.2 J (D) need more information

Q.39

The net charge of the asymmetric object is – (A) Positive (B) Negative (C) Zero (D) Cannot determine

CAREER POINT __________________________________________________________Electrostatics

a(xî  yĵ  zk̂ ) , where a = 122.5 (x 2  y 2  z 2 )3 / 2 SI unit and is a constant. Find the potential difference (in volt) between (3, 2, 6) and (0, 3, 4).

Q.44

A thin wire ring with a radius R and mass m carries an electric charge q. The centre of the ring contains a charge Q of the same sign as q and Q >>q. If the ring rotates with an angular velocity about its centre. Find the increment in tension developed in the ring.

Q.45

A thread carrying a charge (uniform)  per unit length has configuration shown in figure. A D

Q.40(a) An infinite number of point charges of equal magnitude q, but of opposite sign consecutively are placed along the x-axis at x = 1 m, x = 2 m, x = 3 m and x = 4 m and so on upto +. What is the value of net electrostatic force at the point, x = 0 where positive charge q0 is present ? (b) The electric field strength depends only on the x, y and z coordinates according to the

(b) A clock face has –ve point charges –q, –2q, –3q, ...., –12q fixed at the positions of the corresponding numerals. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the same direction as the electric field vector at the centre of the dial ?

law E =

Q.41

Two spherical bobs of same mass and radius having equal charges are suspended from the same point by strings of same length. The bobs are immersed in a liquid of relative permittivity r and density o. Find the density  of the bob in g/cm3 for which the angle of divergence of the strings to be the same in the air and in the liquid (r = 3, 0 = 2g/cm3) ?

Part-D : Numerical Response/Subjective Type Qs.

An inclined plane makes an angle of 300 with the horizontal electric field E of 100 V/m. A particle of mass 1 kg and charge 0.01 C slides down from a height of 1 m. If the coefficient of friction is 0.2, find the time taken for the particle to reach the bottom.

EE R

Q.42

Q.46

E = 100V/m

30º

Q.43(a)A particle of mass m carrying charge 'q' is projected with velocity (v) from point P towards an infinite line charge from a distance 'a'. Its speed reduces to zero momentarily at point Q which is at a distance a/2 from the line charge. If another particle with mass m and charge – q is projected with the same velocity v from point P towards the line charge. Its speed is Nv found to be at point 'Q'. Find the value 2 of N.

Q.47

O R C

Assuming a curvature radius R to be considerably less than the length of thread. Find the magnitude of electric field strength at point O.  The angle of E at point P due to uniformly  charged finite rod will be radian, with x a axis then value of a is d x  + 90º P + + + + 30º + Find the force in N experienced by the semicircular rod of radius R charged with a charge q, placed as shown in figure. The line of charge with linear charge density  is passing through its centre and perpendicular to the plane of rod (q = 220 c, R = 1 metre,  = 2C/m) + + q + + + + + + rod R + + + + + + wire + +  +

+ +

Electrostatics _________________________________________________________CAREER POINT

Q.52

A long cylindrical wire carries a positive charge of linear density 2.0 × 10–8 C/m. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. If it is a × 1.44 × 10–17 Joule.

Q.53

A proton approaches with initial velocity 0 an another proton which is kept at infinity at rest initially. Find the closest approach in this case take mp as mass of a proton.

Q.54

Two circular rings A and B, each of radius a = 30 cm are placed coaxially with their axes horizontal in a uniform electric field E = 105 N/C directed vertically upward as shown in figure. Distance between centres of these rings A and B is h = 40 cm. Ring A has a positive charge q1 = 10 µC while ring B has a negative charge of magnitude q2 = 20 µC. A particle of mass m = 100 g and carrying a positive charge q = 10 µC is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance of 40 cm. (Take g = 10 ms–2) if it

vertical wall

q smooth table

The charge Q = C is distributed on a thin semicircular ring of radius R = 2m. There is a uniform electrostatic field | E | = 2N/C directed horizontal. The semicircular ring can rotate freely about a fixed vertical axis AB. Initially the ring is in static equilibrium a shown in figure. If we want to rotate it about the fixed axis by 90º then minimum work required on the ring is xJ. Find the value of x.

EE R

Q.49

A block of mass m containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure. The distance of the block from the wall is d. A horizontal electric field E towards right is switched on. Assuming elastic collisions (if any) find the time period of the resulting oscillatory motion in second. Is it a simple harmonic motion N q    1C / kg, E  1 , d  2m  ? C m  d

Q.48

is v m/s. Find v/ 2 . 

 E

Two small equally charged identical conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is 10 cm (the length of the threads >> 10 cm.) One of the balls is then discharged. How will the balls behave after this ? What will be the distance between the balls when equilibrium is restored ?

Q.50

Q.55

In the figure shown S is a large nonconducting sheet of uniform charge density A rod R of length  and mass ‘m’ is parallel to the sheet and hinged at its mid point. The linear charge densities on the upper and lower half of the rod are shown in the figure. Find the angular acceleration of the rod just after it is a released. If it is . Find a. 2m 0 

 Three charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the  hinged point positions where the charges should be placed such that the potential energy of – this system is minimum. CAREER POINT __________________________________________________________Electrostatics

Q.51

Q.60



Q.61 Q.57

Charges +q and –q are placed at the corners of a cube of side a as shown in the figure. Find the work done to separate the charges to infinite distance

EE R

Q.58

The electric potential in a region is given by V(x, y, z) = ax2 + ay2 + abz2, 'a' is a positive constant of appropriate dimensions and b, a positive constant such that V is volts when x, y, z are in m. Let b = 2. The work done by the electric field when a point charge + 4C moves from the point (0, 0, 0, 1m) to the origin is 50 J. The radius of the circle of the equipotential curve corresponding to 2 m is  m. Find

V = 6250 volts and z = 2.

Q.59

A particle having mass m and charge (–q) moves along an ellipse around a fixed charge Q such that its maximum and minimum distances from the fixed charge are r1 and r2 respectively. Show that the angular momentum L of this particle about Q is a × 104 Ns, then find a. (Q = 2C, q = 1C, r1 = 3m, r2 = 1m, m = 0.3kg) r2 +Q

–q

Small identical balls with equal charges are fixed at vertices of regular polygon with side a. At a certain instant, one of the balls is released & a sufficiently long time interval later, the ball adjacent to the first released ball is freed. The kinetic energies of the released balls are found to differ by K at a sufficiently long distance from the polygon. Determine the charge q of each 1 × 10–4 C. Find a. (k = 10 ball. If it is a Joule, side length = 1m)

– 3 / 2 m, + 3 / 2 m and + 27 / 2 m respectively on the y-axis. A particle of mass 6 × 10–4 kg and of charge +0.1µC moves along the negative x direction. Its speed at x = +  is v0. Find the least value of v0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given 1/(40) = 9 × 109 Nm2/C2.

S 

Four point charges + 8µC, –1 µC, –1µC and + 8µC are fixed at the points – 27 / 2 m,

A simple pendulum of length  and bob mass m is hanging in front of a large nonconducting sheet having surface charge density . If suddenly a charge +q is given to the bob & it is released from the position shown in figure. Find the maximum angle through which the string is deflected from vertical.

Q.56

–q

+q +q

Q.62

Charge +Q fixed at the origin of the coordinate system. A small electric dipole of dipole moment P is pointing away from the charge along the x-axis. The dipole is released from a point far away from the origin. (i) Find the kinetic energy of the dipole when it reaches to a point (d, 0). (ii) At this moment find the force acting on charge +Q.

Q.63

Two uniformly charged large plane sheets S1 and S2 having charge densities 1 and 2 (1 > 2) are placed at a distance d parallel to each other. A charge q0 is moved along a line of length a(a < d) at an angle 45º with the normal to S1. Calculate the work done by the electric field, in joule (q0 = 2C, 1 = 30 C/m2, 2 = 0 C/m2, a = 2 metre).

(–q)

Electrostatics _________________________________________________________CAREER POINT

EXERCISE (Level-4) Old Examination Questions

Q.1

Q.4

An electric dipole is placed at an angle of 30º to a non-uniform electric field. The dipole will experience – [AIEEE 2006] (A) a torque as well as a translational force (B) a torque only (C) a translational force only in the direction of the field (D) a translational force only in a direction normal to the direction of the field

Q.5

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 – V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is relaeased from rest on the inner surface of plate 1. What is its speed when it hits plate 2 ? (e = 1.6 × 10–19 C, me = 9.11 × 10–31 kg) – [AIEEE 2006]

A charged ball B hangs from a silk thread S which makes an angle  with a large charged conducting sheet P, as shown in

the figure. The surface charge density  of

+ + P + +  S + + + + B

Q.2

(A) cos 

(B) cot 

(D) tan 

the sheet is proportional to- [AIEEE-2005]

Section-A [JEE Main]

Two point charges +8q and –2q are located

The

EE R

at x = 0 and x = L respectively.

0.1 m

location of a point on the x axis at which

the net electric field due to these two point [AIEEE-2005]

charges is zero is (A) 2 L (C) 8 L

(A) 1.87 × 106 m/s (C) 2.65 × 106 m/s

(D) 4 L

Two thin wire rings each having a radius R

Q.3

(B) L/4

Q.6

are placed at a distance d apart with their axes coinciding.

The charges on the two

rings are +q and –q.

difference between the centres of the two [AIEEE-2005]

rings is -

(A)

QR/40d2

(B)

Q 2 0

1 1    R R2  d 2

  

1 1   2  R R  d2

  

Q 4 0

Q.7

(B) 32 × 10–19 m/s (D) 7.02 × 1012 m/s

An electric charge 10–3 C is placed at the origin (0,0) of X - Y co-ordinate system. Two points A and B are situated at ( 2 , 2 ) and (2, potential difference and B will be – (A) 9 volt (C) 2 volt

The potential

0) respectively. The between the points A [AIEEE-2007] (B) zero (D) 4.5 volt

Charges are placed on the vertices of a  square as shown. Let E be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then [AIEEE 2007]

CAREER POINT __________________________________________________________Electrostatics

–q D

Two points P and Q are maintained at the potentials of 10 V and – 4V, respectively. The work done in moving 100 electrons from P to Q is – [AIEEE-2009] (A) –9.60 × 10–17 J (B) 9.60 × 10–17 J (C) –2.24 × 10–16 J (D) 2.24 × 10–16 J

Q.12

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals – [AIEEE-2009] (A) – 2 2 (B) – 1 1 (C) 1 (D) – 2

Q.13

A thin semi-circular ring of radius r has a positive charge q distributed uniformly

C –q

 (A) E remains unchanged, V changes  (B) Both E and V change  (C) E and V remain unchanged  (D) E changes, V remains unchanged Q.8

Q.11



over it. The net field E at the centre O is – [AIEEE-2010]

The potential at a point x (measured in m) due to some changes situated on the xaxis is given by V (x) = 20 /(x2 – 4) volts. The electric field E at x = 4 m is given by [AIEEE 2007] (A) 5/3 Volt/m and in the –ve x direction (B) 5/3 Voltm and in the +ve x direction (C) 10/9 Volt/m and in the –ve x direction (D) 10/9 Volt/m and in the +ve x direction

Q.9

EE R

If gE and gm are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the moon/electronic charge on the earth) to be [AIEEE 2007] (A) 1 (B) 0 (C) gE/gM (D) gM/gE

Statement-1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement-2 : The net work done by a conservative force on an object moving along a closed loop is zero. [AIEEE-2009] (A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is true. Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (C) Statement-1 is true, Statement-2 is false. (D) Statement-1 is false, Statement-2 is true.

(A)

CA |

q 2

2  0 r

4  0 r

î

(B)

ĵ

q 2

ĵ

q 4  0 r 2

(D) 

ĵ

q 2

2  0 r 2

ĵ

Q.14

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is [AIEEE-2010] (A) 1 (B) 4 (C) 3 (D) 2

Q.15

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d<<l) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them [AIEEE-2011] (A) v  x–1/2 (B) v  x–1 (C) v  x1/2 (D) v  x

Q.10

Electrostatics _________________________________________________________CAREER POINT

1 2qQ 4 0 a

 2  1 –    5 

(D)

1 2qQ 4 0 a

 1  1 –    5 

EE R

(B)

A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is [JEE-Main 2013] A

B L

Q.19

(A)

Q 4 0 L ln 2

(B)

Q ln 2 4 0 L

(C)

Q 8 0 L

(D)

3Q 4 0 L

Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of x



mass m and charge q0 = | t | dt, is placed 0

[JEE Main Online 2013]

kmg tan 

kmg / tan 

(B)

(D) 4

kmg tan  kmg tan 

Q.21

A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.5 kV m–1. the density of liquid is 1.26 × 103 kg m–3. The radius of the drop is (neglect buoyancy) – [JEE Main Online 2013] (A) 4.3 × 10–7 m (B) 7.8 × 10–7 m –7 (C) 0.078 × 10 m (D) 3.4 × 10–7 m

Q.22

Statement-1 : No work is required to be done to move a test charge between any two points on an equipotential surface. Statement-2 : Electric lines of force at the equipotential surfaces are mutually perpendicular to each other [JEE Main Online 2013] (A) Statement-1 is true, Statement 2 is true, Statement-2 is the correct explanation of Statement-1 (B) Statement-1 is true, Statement 2 is true, Statement-2 is not correct explanation of Statement-1 (C) Statement-1 is true, Statement-2 is false (D) Statement-1 is false, Statement-2 is true

(D)

(C)

Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating massless strings of equal lengths. They come to equilibrium with each string making angle  from the vertical. If the mass of each charge is m, then the electrostatic potential at the centre of line joining them will be  1     4   k  . 0  

Three positive charges of equal value q are placed at vertices of an equilateral triangle. The resulting lines of force should be sketched as in%[JEE Main On line 2012]

(A)

Q.18

Q.20

(C)

displacement (y << a) along the y-axis, the net force acting on the particle is proportional to [JEE-Main 2013] 1 1 (A) (B) – (C) y (D) – y y y

Q.17

Two positive charges of magnitude 'q' are placed at the ends of a side (side 1) of a square of side '2a'. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charges Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is [AIEEE-2011] (A) zero 1 1  2qQ  1   (B)  4 0 a  5

Q.16

at the origin. If charge q0 is given a small CAREER POINT __________________________________________________________Electrostatics

Q.23

The surface charge density of a thin charged disc of radius R is . The value of the electric field at the centre of the disc is  . With respect to the field at the 2 0

Q.28

2a 2Q

(A) (C)

1 6p1 p 2 4 0 x 2

Two balls of same mass and carrying equal charge are hung from a fixed support of

 Assume that an electric field E  30x 2 î

Q.27

exists in space. Then the potential difference VA – VO, where VO is the potential at the original and VA the potential at x = 2 m is - [JEE-Main 2014] (A) –120 V (B) – 80 V (C) 80 V (D) 120 V

2( b  a 2 ) 2Q

a 2

Within a spherical charge distribution of charge density (r), N equipotential surfaces of potential V0, V0 + V, V0 + 2V, ……V0 + NV (V > 0), are drawn and have increasing radii r0, r1, r2, ……. rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and V then : [JEE-Main Online - 2016] 1 (A) (r)  2 (B) (r)  r r 1 (C)(r)  (D) (r) = constant r

1 6p1 p 2 4 0 x 4

length . At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, x between the balls is proportional to : [JEE Main Online 2013] (A)  (B) 2 (C) 2/3 (D) 1/3

(D)

Q 2

Q.30

[JEE Main Online 2013] 1 3p1 p 2 (B) 4 0 x 3 (D)

(B)

The potential (in volts) of a charge distribution is given by V(z) = 30 – 5z2 for |z|  1 m V(z) = 35 – 10 |z| for |z|  1 m V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume 0 (in units of 0) which is spread over a certain region, then choose the correct statement. [JEE-Main Online - 2016] (A) 0 = 10 0 for |z|  1 m and 0 = 0 elsewhere (B) 0 = 20 0 in the entire region (C) 0 = 40 0 in the entire region (D) 0 = 20 0 for |z|  1 m and  0 = 0 elsewhere

EE R

1 4 p1 p 2 (A) 4 0 x 4

(a 2  b 2 )

Q.29

 Two point dipoles of dipole moment p1 and  p2 are at a distance x from each other and   p1 || p2 . The force between the dipoles is-

Q.25

 A uniform electric field E exists between the plates of a charged condenser. A charged particle enters the space between  the plates and perpendicular to E . The path of the particle between the plates is a[JEE Main Online 2013] (A) straight line (B) hyperbola (C) parabola (D) circle

Q.24

centre, the electric field along the axis at a distance R from the centre of the disc. [JEE Main Online 2013] (A) reduces by 70.7% (B) reduces by 29.3% (C) reduces by 9.7 % (D) reduces by 14.6%

The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has A volume charge density  = , where A is a r constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is [JEE-Main 2016]

Electrostatics _________________________________________________________CAREER POINT

Q.32

Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities + , –  and +  respectively. The potential of shell B is : [JEE Main - 2018]     a 2 – b2   a 2 – b2 (A)  c  (B)  c    0  a  0  b    b2 – c 2   a   b 

(D)

 0

 b2 – c 2   a   c 

Section-B [JEE Advanced]

Q.1

A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct ? [JEE-Main Online-2018] (A) The total energy of the system is 1 1 q 2 E2 m2 A 2  2 2 k (B) The new equilibrium position is at a 2qE distance : from x = 0 k (C) The new equilibrium position is at a qE distance: from x = 0 2k (D) The total energy of the system is 1 1 q 2E2 m2 A 2  2 2 k

R CA Q.34

A solid ball of radius R has a change density r   given by  = 0 1   for 0  r  R. The R  electric field outside the ball is [JEE-Main Online-2018] (A)

0 R 3

(C)

30 R 3

0r 2 4 0 r 2

(B)

40 R 3

(D)

0 R 3

3 0 r 2

Three large charged sheets having surface charge density as shown in the figure. The sheets are placed parallel to XY plane. Then electric field at point P [IIT-JEE 2005]

Z = 3a

K̂

 P

–2

Z=a

EE R

Q.33

 0

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to [JEE-Main Online-2018] 3F F (A) (B) 4 2 3F (C) F (D) 8

(C)

Q.35

An electric dipole has fixed dipole moment  p , which makes angle  with respect to x-axis . When subjected to an electric field   E1 = E î , it experience a torque T1 = k̂ . When subjected to another electric field    E 2 = 3 E1 ĵ it experiences a torque T2 = – T1 . The angle  is. [JEE-Main - 2017] (A) 30° (B) 45° (C) 60° (D) 90°

Q.31

–

Z=0

Q.2

(A)

4 ^ k 0

(B)

4 ^ k 0

(C)

2 ^ k 0

(D) 

2 ^ k 0

Positive and negative point charges of a  equal magnitude are kept at  0, 0,  and 2  –a   0, 0,  , respectively. The work done by 2   the electric field when another positive point charge is moved from (–a, 0, 0) to (0, a, 0) is [IIT-JEE 2007] (A) positive (B) negative (C) zero (D) depends on the path connecting the initial and final positions

12 0 r 2

CAREER POINT __________________________________________________________Electrostatics

Q.3

Consider a system of three charges

(A) |Q1| > |Q2| (B) |Q1| < |Q2| (C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero

q q , 3 3

2q placed at points A, B and C, 3 respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60°. Figure [IIT-JEE 2008]

and –

Q.6

C O

B x

60° A

(A) The electric field at point O is

8 0 R 2 directed along the negative x-axis (B) The potential energy of the system is zero (C) The magnitude of the force between the

EE R

54  0 R 2 q (D) The potential at point O is 12 0 R

Q.4

Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbital. Find out the correct statement(s). [IIT-JEE 2009] (A) The angular momentum of the charge – q is constant (B) The linear momentum of the charge –q is constant (C) The angular velocity of the charge –q is constant (D) The linear speed of the charge –q is constant

Q.5

A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that MCQ [IIT-JEE 2010]

(A) of the same frequency and with shifted mean position. (B) of the same frequency and with the same mean position (C) of changed frequency and with shifted mean position (D) of changed frequency and with the same mean position

charges at C and B is

A wooden block performs SHM on a frictionless surface with frequency, v0. The block carries a charge + Q onits surface. If now a uniform electric field E is switchedon as shown, then the SHM of the block will be [IIT-JEE 2011]  E +Q

Q.7

Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that q 1 K = , which of the following 40 L2 statement(s) is (are) correct ? [ MCQ [IIT-JEE 2012] +q

P E3 A +2q

T O

D –2q

E2 R +q B

E –q

C –q

(A) The electric field at O is 6K along OD. (B) The potential at O is zero (C) The potential at all points on the line PR is same (D) The potential at all points on the line ST is same

Electrostatics _________________________________________________________CAREER POINT

List I

(Q) Q1, Q2, positive; Q3, Q4 negative

(2)

–x

(R) Q1, Q4, positive; Q2, Q3 negative

(3)

(S)

(4)

Q1, Q3, positive; Q2, Q4 negative R-4, R-3, R-2, R-1,

 2 (C) E1 (r0 / 2) = 2E2 (r0 /2) (D) E2 (r0 /2) = 4E3 (r0 /2)

(B) r0 =

–y

Let E1(R) and E2 (r) and E3 (r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density , and an infinite plane with uniform surface charge density  . If E1 (r0) = E2 (r0) = E3 (r0) at a given distance r0, then [IIT-JEE 2014]

(A) Q = 4r02



v –q

(A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge –q continues moving in the direction of its displacement. (D) Charge –q executes simple harmonic motion while charge +q continues moving in the direction of its displacement.

S-2 S-1 S-4 S-3

Q-1, Q-2, Q-1, Q-2,



EE R

(1)

The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density  are kept parallel to each other. In their resulting electric field, point charges q and –q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statements(s) is(are) [JEE Advance - 2015]



List II

Q1, Q2, Q3, Q4 all positive

Code : (A) P-3, (B) P-4, (C) P-3, (D) P-4,

Q.9

(+a, 0) (+2a, 0)

(–2a, 0) (–a, 0)

(P)

Q.10

Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x-axis at x = –2a, –a, +a and +2a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists. [IIT-JEE 2014] q (0, b)

Q.8

Q.11

A particle of mass 10–3 kg and charge 1.0 C, is initially at rest. At time t = 0, the particle comes under the influence of an electric field E (t) = E0 sin t î , where E0 = 1.0 N C–1 and  = 103 rad s–1. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms–1, attained by the particle at subsequent times is ____________. [JEE-Advanced-2018]

CAREER POINT __________________________________________________________Electrostatics

Q.12

LIST-I E is independent of d 1 E d

LIST-II A point charge Q at the origin

EE R

A small dipole with point charges Q at (0, 0, l) and – Q at (0, 0, –l). Take 2l<< d R. 3. An infinite line 1 E 2 charge coincident d with the x-axis, with uniform linear charge density  S. 4. Two infinite wires 1 E 3 carrying uniform d linear charge density parallel to the x-axis. The one along (y = 0, z = l) has a charge density + and the one along (y = 0, z = –l) has a charge density –. Take 2l << d 5. Infinite plane charge coincident with the xy-plane with uniform surface charge density (A) P  5; Q  3, 4; R  1; S  2 (B) P  5; Q  3; R  1, 4; S  2 (C) P  5; Q  3; R  1, 2; S  4 (D) P  4; Q  2, 3; R  1; S  5

The electric field E is measured at a point P(0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distribution. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II [JEE-Advanced-2018]

Electrostatics _________________________________________________________CAREER POINT

EXERCISE (Level-5) Review Exercise y + + + + +

Q.5



P2 x

(A) they will repel each other (B) they will attract each other (C) force of interaction is of magnitude of 3P1 P2

Q.6

EE R

4 0 x 4

Two small spheres having same mass and charge are located at same vertical line at heights h1 and h2 (h2 > h1). The both are thrown simultaneously with the same velocity v in the same direction along the horizontal. From the initial vertical line, the first sphere hits the ground at a distance . Find at this instance height of the second ball, H. A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the energy required to take a test charge q from infinity to apex A of cone. The slant length is L.

(D) force of interaction is of magnitude of 6P1 P2

+ + + + + + ++ A ++ + + + AB = L +++ B+

4 0 x 4

Q.3

– –– ––

––

Figure shows two short dipole moments parallel to each other and placed at a distance x apart is, then –  P1

++ ++

Q.2

A small ball of mass 2 × 10–3 kg having a charge of 1C is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution (A) 1.86 m/s (B) 5.86 m/s (C) 10 m/s (D) 20 m/s

Q.1

Two positive charges q1 and q2 are located 

at the points with radius vectors r1 and 

r2 . Find a negative charge q3 and a radius 

vector r3 of the point at which it has to be placed for the force acting on each of the three charges to be equal to zero.

Q.4

A non-conducting ring of mass m and radius R is charged as shown. The charged density i.e. charge per unit length is . It is then placed on a rough non-conducting horizontal surface plane. At time t = 0, a  uniform electric field E  E 0 î is switched on and the ring start rolling without sliding. Determine the friction force (magnitude and direction) acting on the ring, when it starts moving.

Q.7

Two fixed equal positive charges, each of magnitude q = 5 × 10–5C are located at points A and B separated by a distance of 6m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C. [IIT-JEE 1985] +q

–q C

+q B

CAREER POINT __________________________________________________________Electrostatics

Q.9

A shower of protons from outer space deposits equal charges +q on the earth and the moon, and the electrostatic repulsion then exactly counter balances the gravitational attraction. How large is q ? [REE- 99]

Q.13

A particle of charge q and mass m moves rectilinearly under the action of an electric field E = A – Bx where B is a + ive constant and x is a distance from the point where the particle was initially at rest. Calculate: (i) distance travelled by the particle till it comes to rest and (ii) acceleration at that moment. [REE - 99]

A circular ring of radius R with uniform positive charge density  per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the

Q.12

Two fixed charges –2Q and Q are located at the points with coordinates (–3a,0) and (+3a, 0) respectively in the x-y plane. (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Give the expression V (x) at a general point on the x-axis and sketch the function V (x) on the whole x-axis. (c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so. [IIT-JEE 91]

Q.8

EE R

point P ( 3 R, 0, 0) on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P. [IIT-JEE 93] Q.10

Three particles, each of mass 1g and carrying a charge q, are suspended from a common point by insulated massless strings, each 100 cm long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side length 3 cm, calculate the charge q on each particle. (Take g = 10m/s2). [IIT -JEE 1998]

Q.11

A non-conducting disc of radius a and uniform positive surface charge density  is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has q/m = 4 0g/.

(a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. [IIT-JEE 99]

Electrostatics _________________________________________________________CAREER POINT

ANSWER KEY EXERCISE (Level-1) 2. (C) 9. (D) 16. (B) 23. (B) 30. (B)

3. (B) 10. (B) 17. (B) 24. (C)

4. (B) 11. (B) 18. (A) 25. (D)

5. (A) 12. (A) 19. (B) 26. (B)

EXERCISE (Level-2) 3. (B) 10. (A) 17. (B) 24. (B) 31. (D) 38. (D) 45. (A) 52. (B)

4. (D) 11. (B) 18. (A) 25. (C) 32. (A) 39. (D) 46. (B) 53. (C)

5. (A) 12. (B) 19. (A) 26. (C) 33. (A) 40. (C) 47. (C) 54. (B)

7. (D) 14. (A) 21. (B) 28. (A)

6. (A) 13. (A) 20. (B) 27. (D) 34. (A) 41. (B) 48. (A) 55. (C)

7. (A) 14. (D) 21. (B) 28. (A) 35. (C) 42. (D) 49. (A) 56. (C)

5. (B) 12. (A,B,C,D) 19. (A,B,C)

6. (B,C) 13. (A,B,D) 20. (A,D)

7. (A,B,C,D) 14. (A,B,C) 21. (A)

29. (D) 36. (C)

30. (B) 37. (A)

31. (D) 38. (B)

2. (B) 9. (B) 16. (D) 23. (B) 30. (A) 37. (B) 44. (B) 51. (A)

1. (D) 8. (C) 15. (D) 22. (C) 29. (B) 36. (C) 43. (D) 50. (B)

6. (D) 13. (A) 20. (B) 27. (B)

1. (C) 8. (B) 15. (B) 22. (D) 29. (C)

EXERCISE (Level-3) Part-A

2. (A,C) 9. (A,C) 16. (C) 23. (A)

3. (A,B,C) 10. (A,D) 17. (D)

4. (A,B,C) 11. (B,D) 18. (B)

EE R

1. (A,B,C) 8. (A,C) 15. (A,B,D) 22. (A)

Part-B

24. A  P, R, S ; B  P, Q, S ; C  P, Q, S ; D  P, R, S

Part-C

26. (C) 33. (D)

25. (C) 32. (D) 39. (A)

qq 0 log e 2 N (b) 7 40. (a) 4 0

27. (B) 34. (D)

28. (B) 35. (C)

Part-D

41. 3

42. t = 1.34 s

43. (a) 2, (b) 9.30

 4  0 m 2 R 3  Qq  44. T  46. 6 47. 4  45. Zero 8 2 0 R 2   50. [10(1/4)1/3 cm] 51. q should be placed at 3 cm from 2q

  e2  53.    0m p v 0 2 

54. 6

58. 6

60. vmin = 3 m/s, K.E.at origin = 3 × 10–4 J

61. –

59. 9

1 q2 4 . [3 3  3 6  2 ] 4 0 a 6

55. 3

62. (i)

48. 4 52. 2

 q   56. 2 tan –1    20mg 

P Q QP ; (ii) 4 0 d 2 2 0 d 3

49. 8

57. 3

along positive x-axis

63. 2 CAREER POINT __________________________________________________________Electrostatics

EXERCISE (Level-4) SECTION-A 1. (D) 8. (D) 15. (A) 22. (C) 29. (A)

2. (A) 9. (A) 16. (D) 23. (A) 30. (C)

3. (B) 10. (A) 17. (A) 24. (C) 31. (C)

1. (D) 8. (A)

2. (C) 9. (C)

3. (C) 10. (C)

4. (A) 11. (D) 18. (B) 25. (B) 32. (B)

5. (C) 12. (A) 19. (D) 26. (D) 33. (A)

6. (B) 13. (D) 20. (D) 27. (B) 34. (D)

5. (A,D) 12. (B)

6. (A)

7. (A,B,C)

4. (A) 11. 2.00

SECTION-B

7. (D) 14. (D) 21. (B) 28. (A) 35. (D)

EXERCISE (Level-5) 2. A,C

 5. h1 + h2 – g   V

3. q3 =

Qq 2 0 L

8. (a) radius = 4 a, centre at (5a, 0) ;

( q1  q 2 )2



and r3 =



q 2 r1  q1 r2 q1  q 2

1. B

q1q 2

7. Maximum distance from O = 8.48 m  1 2  (b) V = Q/40    | ( x  3a ) | | (x  3a ) |

q 2 0 m

EE R

10. 3.17 × 10–9 C

12. q = 4 0 GM e M m

4. RE0 î

13. x = 0, x =

11. (a) H =

Qq 8 0 ma

4a a (b) z0 = 3 3

2A qA ,a=– B m

All exercise's detailed solutions are available at www.careerpoint.ac.in

Electrostatics _________________________________________________________CAREER POINT