Chapter 30 by John Smith

POTENTIAL AND FIELD

30.1. Solve: The potential difference AV between two points in space is 9

AV = V(xf) - V(x,) = - I E , dx x,

where x is the position along a line from point i to point f. When the electric field is uniform, xr

AV = - E x j d . =-E,&

= -(lo00 V / m)(0.30 m - 0.10 m) = -200 V

30.2. Visualize:

Solve: The potential difference AV between two points on the y-axis is

-7

E, dy x When the electric field is uniform, the above result simplifies to AV = -E, Ay. In the present problem, AV =

Ay = yr - yj = 0.05 m - (-0.05 m) = 0.10 m The ycomponent of the electric field is E, = -50,000 V / m 3 AV= - (-50,000 V/m)(O.lO m) = +5000 V Assess: V, - V, = 5000 V shows that the potential at point f is higher than at point i. This is because the electric field is directed from the point at the higher potential to the point at the lower potential.

30-1

_._

30-2

Chapter 30

30.3. Visualize:

v = -1Ooov -

v = +loooV + 1.0

-1.0

x (m)

f~--------------------------------i

Solve: The electric potential difference AV between two points in a uniform electric field is given by Equation 30.4: V(xJ - V(XJ = -Ex (xf- x,) Choosing x, = -1 .O m and x, = +1 .O m,

+lo00 V - (-1000 V)=-E,[l.O m-(-1.0

m ) ] a E,=-lOOO Vlrn

Alternatively, x, = 1 .O m and x, = -1 .O m. For this choice, -1000 V - (+lo00 V) = -Ex [-1.0 m - (1.0 m)]

* Ex=-1000

Vlrn

Assess: The choice of initial and final positions does not change the physical nature of the electric field or the potential difference.

30.4. Model: The electric field is the negative of the derivative of the potential function. Solve: From Equation 30.8, the component of the electric field in the s-direction is E, = - d V / d s . For the given potential, d dV = -(loox'

V v) = 200x -

E, = -200x

A t x = Om, E, = 0 Vlm ,and at x = 1 m, E, =-200 (1) V/m=-200 Vlrn. Assess: The potential increases with x, so the electric field must point in the -x direction.

30.5. Model: The electric field is the negative of the slope of the graph of the potential function. Solve: There are three regions of different slope. For 0 cm < x < 10 cm and 20 cm < x < 30 cm, AV -=OVlm*

E,=OVIm

For lOcm < x < ?Ocm,

AV - -100 _ Ax

V - (100 V) = -2000 V I m 0.20m-0.lOm

E, = +2000 Vlrn

30 I Because E, = - d V / d s , the electric field is zero where the potential is not changing. 10

Assess:

30.6. Model: The electric field is the negative of the slope of the graph of the potential function. Solve: There are three regions of different slope. For 0 cm < x < 1 cm, -=

5 0 v - 0 v = 5 0 0 0 ~ / m~ E ~ = - ~ O O O V / ~

0.01m - Om

For1 c m < x < ? c m , ,

AV dx

-50 V - (50 V) = -10,000 V / m 0.02 m-0.01 m

* E , = 10,000 Vlm

For 2 cm < x < 3 cm,

OV -(-5OV) = 5000 V I m a E,= -5000 Vlm 0.03 m - 0.02 m

Potential and Field

Assess:

30-3

AV/& and E, are the negative of each other.

30.7. Model: The electric field points in the direction of decreasing potential and is perpendicular to the equipotential lines. Visualize: Please refer to Figure Ex30.7. The three equipotential surfaces correspond to potentials of 0 V, 100 V, and 200 V. Solve: The electric field component along a direction of constant potential is E, = -dV/ds = 0 V / m. But, the electric field component perpendicular to the equipotential surface is

The direction of the electric field vector is “downhill,” perpendicular to the equipotential surfaces, and directed to the left. That is, the electric field is 10,000 V/m to the left.

30.8. Model: The electric field is perpendicular to the equipotential lines and points “downhill.” Visualize: Please refer to Figure Ex30.8. Three equipotential surfaces at potentials of -200 V, 0 V, and 200 V are shown. Solve: The electric field component perpendicular to the equipotential surface is 200 v E = -% = --- -20,000 V / m As 0.01 m The electric field vector is in the third quadrant, 45” below the negative x-axis. That is, I?

= (20,000 V / m, 45’ below -x-axis)

30.9. Model: Use Kirchhoff s loop law. Visualize: Please refer to Figure Ex30.9. Solve: For any path that starts and ends at the same point, AVmp= C ( A V ) , = 0 V. Therefore, 1

+ AV34 +AV,, =30 V + 50 V +AV3,-60 V = 0 V* AV3,=-20 V AV,? + That is, the potential at point 4 is less than the potential at point 3. 30.10. Model: Any excess charges on a conductor in electrostatic equilibrium are always located on the surface of the conductor. Outside a charged sphere of radius R and charge Q, the electric field and the electric potential are identical to that of a point charge Q at its center. Solve: For r < 10 crn, E = 0 V/m and for r = 10 cm, E = 9000 V/m. For r 2 10 cm, , 1 E=---=

&E0

( 9 ~ 1 Nm’/C’)(10x10-9 0 ~ C)

r’

9 0 ~ ~ ’ / ~ 2 r-

For r 5 10 cm, V = 900 V. For r 2 10 cm,

v=--=1

4zg0 r

( 9 ~ 1 Nm’/C2)(10x104 0 ~ r

c) -- 9 0 N m 2 / C r

Chapter 30

30-4

E (V/m)

10,000

30.11. Solve: The work done is exactly equal to the increase in the potential energy of the charge. That is, W = Ab' = qAV = q(V, - y ) = (1.0 x lo6 C)(1.5 V) = 1.5 x lo6 J Assess: The work done by the escalator on the charge is stored as electric potential energy of the charge.

30.12. Solve: The Van de Graaff generator or the motor that runs the belt does work in lifting a positive ion (q = e ) against the downward force on the positive charge that is moving up the belt. The work done is

w = A u = ~ A v = ~ ( ~ .vo-x o~vo) ~= ( i . 6 0 ~ 1 0 -c)(i.oxio6 ~~ v)=i . 6 0 ~ 1 0 -J ~ ~ Assess: The work done by the generator in lifting the charge is stored as electric potential energy of the charge.

30.13. Solve: The emf is defined as the work done per unit charge by the charge escalator or the battery. That is,

30.14. Solve: (a) Resistivity depends only on the type of material, not the geometry of the wire. Wires 1 and 2 are made of the same material, so p2 = p I and thus p J p , = 1.OO. (b) The resistance of a wire of length L and radius r is given by Equation 30.19:

Because the two wires have the same resistivity,

30.15. Solve: (a) Using Table 30.1 and Equation 30.19, the resistance is R=---A

~ ( 2 . 5 ~ m)* 1 0 ~

(3.5 x IO-' nm)(o.1 m)

(0.001 m)'

(b) The resistance is

R=-=-=

= 3.5

30.16. Solve: We can identify the material by its resistivity. Starting with the wire's resistance, PL PL m 3 R - lr(0.0004 m)2(1.1 Q) = 5.5x R=-=-*p=-A m2 L 10 m From Table 30.1, we can identify the wire as being made of tungsten.

30.17. Model: Assume the battery is an ideal battery. Solve: Connecting the wire to the battery leads to an electric field inside the wire that is given by Equation 30.16:

* AV,,,,, = L E , , = (0.30 m)(10 V/m) = 3.0 V

E,,, = L

Potential and Field

30-5

30.18. Model: Assume the battery is an ideal battery. Solve: (a) The current in a wire is related to its resistance R and to the potential difference AV,,, between the ends of the wire as

From Equation 30.19, (3.0 Q)z(O.3 x lo9 m)’ R = -PL aL=-= = 30.3 m A P 2.8 x lo-’ Szm (b) For a uniform wire of the same material, R = L . I f the wire in this problem is cut in half, its resistance will decrease to +(3 Q) = 1.5 R . Thus, using I = AV,,/R, we have I = 1.0 A.

30.19. Model: Assume the battery is an ideal battery. Solve: We can find the current I from Equation 30.20 provided we know the resistance R of the gold wire. From Equation 30.19 and Table 30.1, the resistance of the wire is

PL p~ R=-=-= A

(2.4 x lo-’ Qm)(lOO m) = 305.6 i2 z(0.0.5~10-~ m)*

AVwk - 0.70V I = -- -= 2.29 mA R

305.6Q

30.20. Model: Assume that the battery is ideal and that the capacitor is a parallel-plate capacitor, Solve: (a) From Equation 30.24, the capacitance is (8.85 x lo-’’ C2 / N m2)(0.02 m x 0.02 m) C=-= = 7 . 1 ~ 1 0 - ’F~ = 7.1pF d 0.0005 m (b) The charges on the electrodes are

F)(100 V) = f7.1 X IO-’’ C = k0.71 nC

Q = C A V , = k(7.1 x

30.21. Solve: From Equation 30.24, the capacitance is

E0 A d

(100 x lo-’’ F)(0.20x m) = 0.0475 m = 4.75 cm 8.85 x IO-’*C2 / N m2

Assess: The capacitance depends only on the geometry of the capacitor. 30.22. Model: Assume that the battery is ideal and the connecting wires have zero resistance. Solve: The capacitance of a capacitor is the ratio of the charge and the potential difference AVc:

For an ideal battery, & = AVbal= AVc. So,

= 3.0 V.

30.23. Model: Assume that the battery is ideal and the wires connecting the battery with the capacitor have zero resistance. Solve: Using Equation 30.26,

30%.

Model: Assume that the battery is ideal. Solve: Equation 30.26 is

30.25. Solve: According to Equation 30.28, C, = C , + C2+ C, = 6 p.F + 10 pF

+ 16 pF = 32

30-6

Chapter 30

30.26. Solve: According to Equation 30.30,

[hl

1 -+-+-c,

k 3 ) =(&+=+=) 1

-1

=3.Mx1O4 F=3.04pF

30.27. Visualize: Please refer to Figure Ex30.27. Solve: Any two electrodes, regardless of their shape, form a capacitor whose capacitance is defined as C = Q/AV, . The capacitance is

30.28. Model: Assume the battery is ideal. Visualize: Please refer to Figure Ex30.28. Solve: For an ideal battery, the potential difference across the capacitor is the same as the emf of the battery. Thus,

Q E = AV - ,-C

F)(1.5 V) = 15.0 X

Q = CE = (10 x

30.29. Solve: From Equation 30.33, the energy stored in a capacitor is U , = +C(AVc)’

AV, =

E -1 =

2(1*oJ,

= 1414 V

1.0~1F 0~

30.30. Solve: The graph of Figure Ex30.30 shows that the capacitor is fully charged at 3.0 s and that its charge is 200 pC. From t = 0 s to t = 3.0 s, the charge on the capacitor follows the equation Q = (200 pC/3.0 s)t . From Equation 30.32, the energy stored in a capacitor as a function of time is

C’ / s F ) t 2

-=(l.lll~lO-~

0.01-

t (s)

30.31. Solve: Using Equation 30.33, V,, = +Cl(AVc,)2

U,, = +Cz(AV,,)’

Because C2 = +Cl and AVc2 = 2AVc,, we have

30.32. Solve: The energy density is uE = (5.0 x lo-” J)/(0.02 m x 0.02 m x 0.02 m) = 6.25 x lo-’ J/m3. uE = i g 0 E 2

E = ~ = ~ 7 = l 21 6.25 90V x lo-’ / m J / m3)

8.85 x

C’ / N m’

Potential and Field

30-7

30.33. Solve: (a) From Equation 30.33, the energy stored in the charged capacitor is 2 1 ~(0.01 m)’(8.85 x IO-” C2 / N m2)

_ -2 -

OSOXIO-~ m (b) From Equation 30.35, the energy density in the electric field is

( 2 0 0 ~ ) ‘= 1 . 1 1 ~ 1 0 J- ~

30.34. Visualize: Please refer to Figure P30.34. Solve: (a) The electric field points “downhill.” So, point A is at a higher potential than point B. (b) In a region that has a uniform electric field, Equation 30.4 gives the potential difference between two points: A V = - E , ( x f -xi)*V,-VA=-(1000V/m)(0.07m)=-70V That is, the potential at point A is 70 V higher than the potential at point B.

30.35. Solve: (a)

E, W m )

(b) Equation 30.3 gives the potential difference between two points in space:

0.30 rn

+[low$]

1000

V = +-[(0.30 2 4.20m

m)’ - (-0.20 m)‘] V = 6 5 V

Assess: E is positive for negative x but negative for positive x, so the potential defference depends on the square of the positions.

30.36. Solve: (a)

v (V)

(b) Equation 30.3 gives the potential difference between two points in space:

-I E& Xt

[;I

- V = -2500(xz Taking V(x,)= 0 V at x, = 0 m, and replacing x, with simply x, V ( x )= -(2500 2)V. AV = V(xf)- V ( x , )=

= -1(5000x V / m)dx = -5000 1,

- x:) V

30-8

Chapter 30

dV = -d(-2500xZ) V = 5 0 0 0 ~= E, d r d x

30.37. Visualize:

Solve: Equation 30.3 gives the potential difference between two points in space A v = V(rf)- V ( 5 )= -iE,dr * 3

r a

i l r R

= -[lnr]; =--[Inr-lnR]=--In2XE, 2XE0

V(r) - V(R) = -j-dr

R 2mOr

A * V(r) = V, --h2XE0

2XE

r R

Assess: At r = R , ln(R/R) = 0 and V(R) = V, as it must be due to the assumed constant of integration.

30.38. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure P30.38. Solve: We have dV dV=-E& dx For x 2 1 cm, E, = 1000 V/m. Integrating the above expression,

E, =--

V = -(lo00 V/m)x + C Because V = 0 V at x = 0.03 m, C = (1000 V/m)(0.03 m) = 30 V. Thus, V = -(lo00 V/m)x + 30 V

This means V at x = 0.02 m is 10 V and V at x = 0.01 m is 20 V. For x < 1 cm, E, = (10’ V/m)x. Therefore, d V = -(IO’ V/m) x dx. Integrating again,

IdV = -j(lO’ V /

rnk dx+ C =

Because V=20 V a t x = 0.01 m, C = 25 V. At x = 0 m, V = 25 V.

v 07)

V = -(los V / m)-+2

Potential and Field

30-9

30.39. Model: Assume the electrodes form parallel-plate capacitors with a uniform electric field between the plates. Visualize:

E, (v/m)

I.'(v)

-1.4 x 10'

Please refer to Figure P30.39. The three metal electrodes serve as plates for two capacitors. On the middle electrode, half the charge is located on the left face and half on the right face, thus forming two capacitors. Each plate of the two capacitors carries a charge of L50 nC. Solve: (a) In the space 0 cm < x < 1 cm, the electric field points to the left and its magnitude is

E = - v= - = E,

Q A&,

50 x C =1.41x107 V / m (0.02 m)'(8.85 x lo-'' C2 / N m 2 )

In the region 1 cm 5 x I 2 cm, = 0 because in electrostatics the inside of a conductor has no free charge. The electric field in the region 2 cm < x < 3 cm points to the right and has the same magnitude as the electric field in the region 0 cm < x < 1 cm. (b) The potential difference between two points in space with a uniform electric field is

AV = V, - y = E(x, - x i )

E, W/m)

-5v

I I +---D1-14--+-+4---bI--+ I I I

1 I

I I -

-1

1 I

IOV

I I

;

I - - - - - d J I I

! ;

I I(

: I'

I I

-2

1 I

I I

I-I-lt

-3

ov;

-1OVI

I I

x(m)

Solve: (a) The x-component of is E, = -dV/&. From Figure P30.40, the slope of the graph can be measured in five different regions. For x < -3 m and x > 3 m, the slope is 0 V/m. For -3 m < x < -1 m and 1 m < x < 3 m the slope is -5 V/m. For -1 m < x < 1 m the slope is +IO V/m. (b) The potential is independent of y, so the equipotential lines are parallel to the y-axis. The x-values of various equipotentials can be read from Figure P30.40. These give the potential map shown above. (c) The field vectors point in the direction of decreasing potential. Furthermore, the results of part (a) show that the field vectors in the inner region (-1 m < x < 1 m) are twice as long as the field vectors in the outer region. As you can see by comparing the vectors to the graph of the electric field, this information was used to add the field vectors to the potential map.

30- 10

Chapter 30

30.41. Model: The electric field is the negative of the slope of the potential function. Solve: The on-axis potential of a charged disk was obtained in Chapter 29 to be

disk

2m0R2

where the charge on the disk of radius R is Q and the point is a distance z away from the center of the disk. Because E = - dV/dz , we have

Assess: Using binomial expansion with z >> R, we have

That is, the disk behaves like a point charge. This is an expected result.

30.42. Model: Assume the charged rod is a line of charge of length L. Visualize: Please refer to Figure P30.42. Solve: (a) Divide the charged rod into N small segments, each of length Ax and with charge Aq. The segment i located at position x,, contributes a small amount of potential V, at point P

Point P is at a distance x, from the origin. This is done to avoid confusion with xi.The Vi are now summed and the sum is converted to an integral giving

Replacing x, with x , the potential due to a line charge of length L at a distance x from the center is

v = -1n(Q

4Z&,L

x+L/2 x - L/2

(b) Because E, = -dV/dx , 1

Assess: When L = 0 m, E , = Q/47r&,,x2 . This is the electric field of a point charge Q a distance x away from a point charge, as expected.

30.43. Model: The electric field is the negative of the slope of the potential graph. Visualize: Please refer to Figure P30.43. Solve: Since the contours are uniformly spaced along the y-axis above and below the origin, the slope method is the easiest to apply. Point 1 is in the center of a 75 V change (25 V to 100 V) over a distance of 2 cm, so the slope AV/& is 37.5 V/cm or 3750 V/m. Point 2 has the same potential difference in half the distance. Thus the slope at point 2 is 7500 V/m. The magnitudes of the electric fields at points 1 and 2 are 3750 V/m and 7500 V/m. The directions of the electric fields are downward at point 1 and upward at point 2 , that is, from the higher potential to the lower potential. That is,

E, = (3750 V/m,

down)

= (7000 V/m, up)

Potential and Field

30-1 1

30.44. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize:

,200v

\ I

t Scale

\ \ \

\-2oOv

8 cm

Solve: (a) At each point, E is perpendicular to the equipotential lines, points “downhill” toward lower potential, and has strength E = dV/ds = AWAs. We can read AVfrom the contour lines and use the 8-cm-scale at the bottom of the figure to estimate As. Point 1, on the 100 V contour, is between the 200 V and the 0 V contours. The actual distance on rhefigurc between the 200 V and the 0 V contours can be measured to be 4 mm. The length of the 8-cm-scale is = 13- rnm.Thus

as,-

200 v ( ~ ) ( 8 c m ) = 4 c m ~ E=-=5000 1 0.040 m

V/m

From the symmetry of the figure, the field strengths are the same at points 2,4, and 5. That is, E , = E2 = E4 = Es 5000 V/m. The contour lines are closer at point 3, so the field strength E3is larger. The measured spacing between the 100 V and -100 V lines, which bracket the point, is ~ 1 . 5mm. Thus

The field strength at 3 is four times that at the other points. (b) The field vectors are shown on the figure. Each is drawn perpendicular to the equipotential at that point.

30.45. Model: The electric field is the negative of the slope of the graph of the potential function. Solve: The electric potential in a region of space is V = (150 x? - 200 y’) V where x and y are in meters. The x- and y<omponents are dV E, = --dV = -(300 x ) V I m E, = -- = +(400 y ) V / m dr dY At (2.0 m, 2.0 m), E , = -600 Vlm and E, = 800 Vlm. The magnitude and direction of the electric field are

,/=

= d(-600 V / m)’ +(SO0 V / m)’ = lo00 V / m

tan8 = E’ = goo I - - a 8 = 53.1O above the -x axis lE,\ 6 0 0 V i m 3

30- 1 2

Chapter 30

30.46. Model: The field strength is the negative of the slope of the graph of the potential function. Solve: (a)

Equipotentials 25

\\\

50\\

Equipotentials are drawn by connecting points of equal potential. (b) To find first determine the direction perpendicular to the equipotential and pointing toward a lower potential. Call this direction the s-axis. The field strength is E = IdV/dsl= AV/As, where As is the distance between equipotential lines that have potential difference AV. The electric field at point A points straight to the left, toward the 50 V equipotential. The distance between the 50 V line and the 75 V line is one grid spacing, or As = 5 cm. Thus,

As a vector, the field is

= (500 V / m, left). Using the 50 V and 100 V equipotentials to calculate the field at point

B, EB = (50 V)/(0.05 m) = 1000 V I m . Thus E, = (1000 V / m, right). The field at point C points at a 45" angle toward the 50 V equipotential. The distance As between the 50 V and 100 V equipotentials is 5 cm X f i = 7.07 cm, so Ec = (50 V)/(0.0707 m) = 707 V. As a vector, E, = (707 V I m, 45" above straight left). The field at point D is symmetrical with point C, so E, = (707 V / m, 45" below straight left). (c) The electric field vectors are shown in the figure.

30.47. Model: Conductors connected by a conducting wire are at the same potential. Visualize: Q21=OC QI1= 6 nC

Switch

Charge flows from sphere 1 to sphere 2 , when the wire is connected. Once connected, the entire system is one large conductor in electrostatic equilibrium. Consequently, the entire system must be an equipotential. That is, the two spheres must have equal potentials VI = V2after being connected by the wire. Solve: The potential of a sphere is V = Q / ~ X E ~soRthe , final charges Q,, and Qzfon the two spheres must be such that Qif

47râ&#x201A;Ź,R,

47râ&#x201A;Ź,R2

47rEO(2R,)

Q,, = +Qz,

where we used the fact that R2 = 2R,. It is also true that charge must be conserved, so Q i f + Q x = Q i , + QZ, = 6 nC Using Q2,= 2 Q , , we find Q i f + 2Qif = 3Qif = 6 nC = 4 nC. from which we finally get Q,, = 2 nC and

Potential and Field

30-13

30.48. Solve: I

I I I I

I \

I I

1 I

\ \

I I

I I I 1 I I \

I I

\ \ \

, v I

-200

+200

-100v

+1mv

v=ov Because the spheres have equal and opposite charges and they have the same diameter, their potential is zero on a line that bisects the two spheres. On the left and right of this 0 V equipotential line, the equipotential lines show the curvature of the sphere the line is nearer to. This is because the potential nearer to one sphere is dominated by the potential of this sphere. The equipotential lines at 0 V, +lo0 V and k2200 V are shown in the figure.

30.49. Model: The potential inside a conductor is the same as the potential on the surface. Solve: The potential on the surface of the copper sphere must be 500 V. That is,

30.50. Model: Assume the battery is ideal. Solve: (a) The electric field inside the wire is E = AV,,,/L. Attaching the wire to the battery makes AVwk = AVba = 1.5 V . Thus,

'.'

E=---10V/m 0.15 m (b) Using Table 30.1, the current density is J=&=-=

E p

10V/m =6.7x106 A / m Z 1.5~1R 0m ~

The resistance is related to the wire's geometry by

Thus, the wire's diameter is d = 2r = 0.62 111111.

30- 14

Chapter 30

30.51. Model: Assume the battery is ideal. Visualize: Area A

-L-

Solve: The tube's resistance is R = -PL =

(1.5 x lo6 SZm)(0.20 m)

z(( - q2) - z[(o.o015 m)* -(0.0014 m)']

= 0.329 SZ

When connected to the battery, AV,,, = AV,,, = 3.0 V. Thus, the current is

30.52. Model: Assume the battery is ideal. Solve: (a) Because the battery provides a current of 0.50 A and the current is defined as the amount of charge that passes through a cross section per second, the charge lifted by the escalator is 0.50 C/s. (b) The work done by the escalator in lifting charge Q is W = U = QAV = (1.0 C)(1.5 V) = 1.5 J (c) The power output of the charge escalator is the work done by the escalator per unit time:

p=-=-=--Au At

QAv - IAV = (0.50 A)(1.5 V) = 0.75 W At

30.53. Model: Assume the battery is ideal. Visualize: The current supplied by the battery and passing through the wire is I = AV,JR. A graph of current versus time has exactly the same shape as the graph of AVbawith an initial value of Io = (AVb,)dR = (1.5 V)/(3.0 Q) = 0.50 A. The horizontal axis has been changed to seconds. I (A)

7200

Solve: Current is I = dQ/dt. Thus the total charge supplied by the battery is Q = r Z dt = area under the current-versus-time graph 0

= +(7200 s)(0.50 A) = 1800 C

30.54. Model: The charged metal plates form a parallel-plate capacitor. Solve: (a) When the charged plates are connected, the maximum current in the wire can be obtained from I,, = AV,,/R where AV,, is the potential difference across the capacitor just before the wire is connected. AVfor the capacitor is AV,, = Q / C . where C is the capacitance for a parallel plate capacitor. Noting that C = &,A/d and R = pUm'. the maximum current is

(13.5 x 1 0 - ~C ) ~ ( O 12 . I x io-3 m)'

(8.85 x IO-''

, =4.17x105 A

C' / N m')(1.7 x IO-' R m)z(5.0 x lo-' m)-

Potential and Field

30-15

In the above calculations, note that the length of the wire L is the same as the separation of the two plates d. Also, the resisitivity of copper is taken from Table 30.1. (b) The largest electric field in the wire is

E,,=-

m ‘a‘

L Calculating AV-

separately,

1800 V ~Em,=-=l.8x105 0.010 m

V/m

the capacitor and hence the current goes down. (d) The energy dissipated in the wire is the energy present in the capacitor before it was connected with the wire. This energy is 1 - ” x Umsrrpaied - -= -Q e = 1QAV,_ = -(12.5 2 2 c 2 c 2

C)(1800 V) = 1.12 x 10” J

30.55. Model: The battery is assumed to be ideal. Visualize: Battery

df=2mm

-1 mm-

di +

The pictorial representation shows the capacitor plates connected to a battery, the battery removed from the plates, and the plates moved apart with insulating handles. Solve: (a) The battery charges the plates through the wires. Once the wires are disconnected, the charge is trapped on the plates and will not change as their spacing is increased. The initial capacitance of the plates is €,A c =-=

’

(8.85 x lo-” C2 / N m2)(0.02m x 0.02 m)

0.001 m

= 3.54 x

Consequently, the initial voltage AV, = 9 V charges the plates to

F)(9 V) = f3.19 x lo-’’ C

Q = fCAV = k(3.54 x

(b) The charge is still Q = S . 1 9 x lo-’’ C after the spacing is increased, but the final capacitance is

c - E A -EoA-C, 0

d, As a result, the final potential difference is

2d,

30-16

Chapter 30

/ A

The pictorial representation shows the capacitor plates connected to a battery, and the capacitor plates moved apart with insulating handles while they are connected to the battery. Solve: (a) The initial capacitance of the plates is

&,A =-=

(8.85 x lo-'' C2 / N m2)(0.02m)*

= 3.54 x lo-'' F

1.0 x 1 0 - ~m

Consequently, an initial voltage AK = 9.0 V charges the plates to Q = +C,AY = k(3.54 x lo-'' F)(9.0 V) = f31.9 x lo-'' C = f31.9 pC (b) The new capacitance is C, = ~,A/2d,= +C,. The potential difference across the plates is determined by the battery and is unchanged: AVf = AK = 9.0. Thus, the new charge on the plates is

Q = kC,AV, = f i ( 3 . 5 4 x lo-'' F)(9.0 V) = 16.0 x lo-" C = 16.0 pC

30.57. Model: Capacitance is a geometric property of two electrodes. Visualize:

1.0 mm

Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q/AV,. The potential difference across the capacitor is

9 RlR2 = (100 x lo-"

F)(1.0 x

m)(9.0 x lo9 N m2 / C') = 900 x 10" m'

Using Rz = R, + 1.O mm,

R , ( R , + ~ . O X ~ Om-)~= 9 0 0 x 1 O 4 m * R : + ( 1 . 0 ~ 1 0 - ~ m ) R , - 9 0 0 ~ 1 0 " m = O

R, =

-1.0 x

m k J(1.0 x

m)'

+ 3600 x 10"

m2 = 0.0295 m = 2.95

The outer radius is R1 = R, + 0.001 m = 0.0305 m = 3.05 cm. So, the diameters are 5.90 cm and 6.10 cm.

30.58. Model: Two capacitors in parallel combine to give greater capacitance. Solve: Since we want a capacitance of 50 pF and we have a 30 pF capacitor, we must connect the second capacitor in parallel with the 30 pF capacitor. That is, C + 30

= 50 @

C = 50 @ - 30 PF = 20 pF

Potential and Field

30-17

30.59. Model: Two capacitors in series combine to give less capacitance. Solve: Since we have a 75 pF capacitor and we want a 50 pF capacitance, we must connect the second capacitor in series with the 75 pF capacitor. The capacitance of the second capacitor is calculated as follows: 1 1 +-=I *C=lSOpF 75pF c 5 o p

30.60. Solve: (a) The equivalent capacitance of N capacitors that are in parallel and have capacitances C,, C,, C,,

. . . C,

c,=c,+c*+c,+

...ICN

If the capacitors are identical with each having a capacitance C, then C, = NC. (b) The equivalent capacitance of N capacitors that are in series and have capacitances C,, C,, C3,... C, is 1 1 1 -=-+-+-+ qe'

...- 1 N'

If the capacitors are identical with each capacitor having a capacitance C, then 1-=

c q c

C N

30.61. Visualize:

fj$ic3=*5.. OTIc3 @----+ C, = 30 /.t.F

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C, and C, are in series, their equivalent capacitance C, ,, is 1 -1 ---+-=-

Cq12 C, Then, C,

1 C,

1 +-=-1 20pF 30pF

1 3 Cql2=12pF 12pF

and C, are in parallel. So,

C,= C,,,

+ C, = 12 ,uF+ 25 pF = 37 pF

30.62. Visualize:

c"q The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C, and C, are in parallel, their equivalent capacitance C, is

= C, + C, = 20,UF + 60pF = 80pF

Then, C, I,and C, are in series. So, 1 1 1 +-=C, CeqIz C,

1 I +-= SOpF 1 0 p F

9 SO

-(pF)-'

Ceq= 9pF = 8.9 pF

30-18

Chapter 30

30.63. Model: Assume that the battery is ideal. Visualize:

The pictorial representation shows the equivalent capacitance of the three capacitors. Solve: Because C,, C2,and C, are in series, 1 1 1 1 1 1 1 1 -=-+-+-=++= -(pF)-' jC, = 2 pF C, C, C, C, 1 2 p F 4 p F 6 p F 2 A potential difference of AV, = E = 30 V across a capacitor of equivalent capacitance 2 pF produces a charge Q = C,AVc = (2 pF)(30 V) = 60 pC. Because C, is a combination of three series capacitors, Q, = Q, = Q3= 60 pC. We are now able to find the potential difference across each capacitor:

Assess:

+ AV, + AK = 30 V = AV,,,

as it should.

fjJ!j12

30.64. Model: Assume that the battery is ideal. Visualize: 0

CS=2gF

c e q

The pictorial representation shows how to fiid the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C, and C, are in parallel,

, and C, are in series, so 1 -=C,

C~,,=C,+C2=4pF+12pF=16pF

1 1 +-=C,,, C,

1 1 18 +-=-(pF) 1 6 p F 2 p F 32

-1

*Ceq = $ pF

A potential difference of AV, = 9 V across a capacitor of equivalent capacitance

6 pF produces a charge

Q = CqAVc = (8pF)9 V = 16 pC Because C, is a series combination of two capacitors C, across C, is

,, and C,, Q, = Q,

Now, Q,, = 16 pC is the charge on the equivalent capacitor with C, the equivalent capacitor C, ,? is

= 16 pC. The potential difference

= 16 pF. So, the potential difference across

Parallel capacitors C, and C2have the same potential difference as the equivalent capacitor C, ,, so AV, = AV, = 1.OV. Thecharge oneachisgiven byQ=CAV, SOQ,=(4pF)(l.OV)=4.0pCand Q2=(12pF)(1.0V)= 12.0pC. In s u m m a r y , Q, = 4.0 pC, AV, = 1.0 V; Q2= 12.0 pC, AV, = 1.0 V; and Q3= 16.0 pC, AV, = 8.0 V. as it should. Assess: Note that AV3 + AVeq = 9.0 V = AV,,,,, as it should. Also that Q, + Q2= 16.0 pC = Q,

@-fq 071

30.65. Model: Assume the battery is an ideal battery. Visualize:

F p 4 = J r @-

5 FF

C3=6pF

5FF

cq2s

czs

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure.

Potential and Field

30-19

Solve: Because C, and C, are in series,

and C, are in parallel, so C,=C,?,+C1=2.4@+5~=7.4~

. A potential difference of AV, = 9 V across a capacitor of equivalent capacitance 7.4 pF produces a charge Q = C,AVc = (7.4 pF)(9 V) = 66.6 pC Because C, is a parallel combination of C , and C, charges on these two capacitors are

Because Q,

?,,

these capacitors have AK = AV,,,

is due to a series combination of C, and C,,

= AV, = 9 V. Thus the

= Q, = 21.6 pC. This means

30.66. Model: Capacitors in parallel add to a greater capacitance compared to individual capacitances. On the

12fiFfi T

other hand, capacitors in series add to a smaller capacitance compared to individual capacitances. Visualize: 12fiF

12 fiF

llyFT

(a)

(b)

12pFTi z p T

T"""

(dl

Solve: (a) Three capacitors in series: 1 -=-

1 1 3 +- 1 +----=-(pF)-' 1 2 p F 1 2 p F 12pF 12

* C,=4pF

(b) Two capacitors in parallel and the third in series with this parallel combination: C,,2 = 12 pF+ 12 p.F= 24pF 1 C,

1 +-=-1 Cq12 1 2 p F

3-=-

(c)

1 +-- 1 - 1 24pF 1 2 p F 8pF

C,=8@

TWOcapacitors in series and the third in parallel with this series combination: 1 C,,>

-=-

* C,

1 +-=-1 1 2 p F 12pF =

* C,

1 (pF)-' 6

=6@

+ 12 pF = 6 pF + 12 ,LLF= 18 pF

(d) Three capacitors in parallel: C, = 12 pF + 12 pF +12 @ = 36 pF

30.67. Model: Capacitance is a geometric property. Visualize: Please refer to Figure P30.67. Shells R, and R, are a spherical capacitor C. Shells RZ and R, are a spherical capacitor C. These two capacitors are in series.

30-20

Chapter 30

Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q/AV,. The potential differences across the capacitors C and C‘ are

Av, =

1 Q -_--_

4 m 0 R,

=,C=(4?E0)

_ [i,_j2]

1 Q = -4n&0R2 4 z 0

,(; i 2 ) ---

C‘ = ( 4 4 (

$- $1

-I

Because these two capacitors are in series,

Assess: C,, depends on only the inner and outer shells, not on R,.

$Vf

30.68. Model: Assume the battery is ideal. Visualize:

4q3c/2

The circuit in Figure P30.68 has been redrawn to show that the six capacitors are arranged in three parallel combinations, each combination being a series combination of two capacitors. Solve: (a) The equivalent capacitance of the two capacitors in series is C. The equivalent capacitance of the six capacitors is I C . (b) As points a and b are midpoints of identical capacitors, V, = V, = 6.0 V. Therefore, the potential difference between points a and b is zero.

30.69. Model: Capacitance is a geometric property and does not depend on the voltage, charge, or any other external factors. We assume that the capacitors are parallel-plate capacitors.

Visualize: Please refer to Figure P30.69. The two electrodes are equivalent to two capacitors in parallel. Solve: The total capacitance of the two capacitors is

= (8.85 x IO-’’ C’ / N m2)(0.01m x 0.01 m)

l + 1.0 x 10-~m 2.0 x 1 0 - ~ m

= 1.33 x IO-’’ F = 1.33 pF

30.70. Model: Assume the battery is ideal. Visualize: Please refer to Figure P30.70. While the switch is in position A, the capacitors C ? and C , are uncharged. When the switch is placed in position B, the charged capacitor C, is connected to C2 and C,. C2 and C, are connected in series to form an equivalent capacitor C,, L3. Solve: While the switch is in position A, a potential difference of V, = 100 V across C, charges it to

Q, =C,V; = ( 1 5 ~ 1 0 “F)(lOOV)=1500,K

Potential and Field

30-21

When the switch is moved to position B, this initial charge Q, is redistributed. The charge Q: goes on C, and the charge Q,, goes on Ce, 23. The voltage across C,and C, is the same and Q: + Q,,, = Q, = 1500 pC. Combining these two conditions, we get

Qi' 1'

- Q,z c,

1500 PC 'q23

Qq 23

Cq 23

-1

Since cqZ3 = (h + h)= 12 p ~we, can rewrite this equation as 1500pc-Q~23 -~Q,23=667pC~Q,'=Q,-Q,,,=1500pC 15 pF 12 pF

,,,

Having found the charge Qeq it is easy to see that Q, = Q3 = 667 pC because C, and C,. Thus,

- 667pC

= 833pC

,,is a series combination of C,

30.71. Model: Assume the battery is an ideal battery. Visualize: Please refer to Figure P30.71. The battery is connected to two series capacitors C, and' C,. Solve: The equivalent capacitance is

Because the charge on capacitor C,is 450 pC, the charge on C, and C, is also 450 pC. We have

Assess: Note that capacitors connected in series have the same charge.

30.72. Model: Assume the battery is ideal. Visualize:

LK--&

= 20 pF

Equilibrium

Connected Disconnected

Solve: When the capacitors are individually charged, their charges are Q, = C, V = (10 pF)( 10 V) = 100 pC

Q2 = C2V = (20 pF)(lO V) = 200 pC

These two capacitors are then connected with the positive plate of C, connected with the negative plate of C? as shown in the figure. Let the new charges on C, and Czbe Q,' and Q;. Then, Q,'+Q;=200pC

- l0OpC

= lOO/fC

The voltages across parallel capacitors are the same, so

Substituting this expression for Q; into the previous equation,

Solving these equations, we get Q,' = 33.3 pC and Q; = 66.7 pC. Finally,

30-22

Chapter 30

30.73. Solve: The magnitude of the work done by the external force is equal to the change in the electric potential energy of the capacitor. The work done is 1 Q' Wfome=AU=Uz-Ul=----2 c,

1 Q2 2 c1

Note that the charge on the plates is not changed as the distance between the electrodes is changed. Thus,

4.0 x

Cy[-

1 2.0pF

5.0p

= 2.4 J

Assess: The work done on the capacitor is stored as electric potential energy in the capacitor.

30.74. Solve: (a) The capacitance of the parallel-plate capacitor is

AEO - (0.1 m x 0.1 m)(8.85~10-" C2 / N m 2 ) = 8.85 x lo-" F =-dl 1.0 x m

The electric potential energy stored in the capacitor is

1 Q' =--=-

2 Cl

1 ( i o ~ i o c)' - ~ = 5.65 x 1 0 - ~J 2 S.85x10-l1F

(b) There is no change in the charge. The energy change is due to the change in the capacitance. The new capacitance is

=o=o=E A E A C, d2

The amount of energy stored is U , = 11.3 x J. (c) Work was done on the capacitor by the agent pulling the plates apart, thereby adding energy into the system.

30.75. Solve: The energy density in the electric field at the surface of the sphere is u = +&,E2,where E is the electric field strength at the surface. The electric field strength and the potential on the surface of the sphere are related as follows:

(' e )

E = - - =I '=-=-4 7 R'~ ~ 4 m 0 R R R =$

looov - 2 . 0 x 1 0 5 N / m

0.005m

u = +(8.85 x lo-'' C2 / Nm2)(2.0 x 10' V / rn)2= 0.177 J / m3

30.76. Model: Conservation of energy. Solve: The energy stored in the capacitor is dissipated through the flash lamp. Power is the rate at which energy is dissipated/absorbed, so the energy dissipated by the flash lamp is PA?=(10W)(lOps)= 1 . 0 1O"J ~

This is the electric potential energy in the capacitor. Using U, = +CAV:, = 1 . 0 ~ 1 J0 ~

+c(~.ov)'

c=22.2 x

30.77. Model: The capacitor is a parallel-plate capacitor and energy is conserved. Solve: The energy needed to melt a 0.50 kg block of ice to water at 0째C is

Q = McAT+ ML, = (0.50 kg)(2090 J / kg K)(10 K ) + (0.50 kg)(3.33 x 10' J / kg) The value of L, is taken from Table 17.3 and the value of c for ice is taken from Table 17.2. Thus, Q = 176,950 J. The capacitor must store electric potential of this amount. Hence,

U c = r.I CAV,)-+C=-(

2Uc - 2(176,950 J) = 141.56 F (AV,)' (50 V)'

Potential and Field

30-23

Because C = A&,I d , the size of the capacitor is A = -c=d E,

(141.56 F)(2.0 x 10-~m) C2 / N m2

8.85 x

= 3.2 x 10" mz

The sheet's edge is = 179 km which is not reasonable.

30.78. Solve: (a) Find an expression for the electric potential in a region where E, = 2z Vim.The potential at z = O is 10 V. (b) Integrating both sides relative to z, V = - 1 2 ~dz+ C = -zz At z = 0 m, V, = 10 V. This means C = 10 V. Thus V = 10 V

- 2.

30.79. Solve:

(a) An iron wire 1.O mm in diameter connects the positive end of a 3.0 V battery to the negative end. If 5.0 A of current is measured in the wire, how long is the wire? (b) The resistance of the wire is

3.0 v 3 R=-=-Q= 5.0 A 5

(9.7 x Q m)L lr(0.00050 m)'

* L=4.86 m

30.80. Solve:

(a) A capacitor is constructed with two 10 cm X 10 cm plates. When the capacitor is connected to a 100 V source, M O O nC is observed on the plates. What is the separation of the plates? (b) From the first equation,

Thus, the second equation becomes 4 . 0 ~ 1 0F ~=

(8.85 x lo-'* Cz / N m2)(0.10m x 0.10 m) d

* d = 2.2 x lo-' m = 0.022 mm

30.81. Solve:

(a) What capacitor should be placed parallel to 3 pF and 6 ,@ capacitors that are in series so that the equivalent capacitance is 4 pF?

(b) We have 18 9

-pF+C=4/fF*C=2pF 30.82. Solve: (a)

-400

The equipotentials are lines of constant potential. The 0 V equipotential requires that v = l o o ~ ~ - y ~ ) = o v ~ ~ ~ = ~ ~ ~ y = ~ These are straight lines at k45" in the xv-plane. The 100 V equipotential is a bit trickier. There,

v = 1O0(x2- y')

= 100 v

* y'

=2 - 12 y =

*Ilx'-1

30-24

Chapter 30

We need to recognize this as the equation for two hyperbolas, one passing through (x, y ) = (1, 0) and the other through (x, y) = (-1, 0). The other equipotentials are also hyperbolas, leading to the potential map shown in the figure. (b) The electric field is

written as a vector, this is

E = -200( xi^- yj) v / m .

30.83. Visualize:

P Solve: The charge density inside the cylinder is p. Consider a Gaussian surface with a radius r inside the cylinder. The charge enclosed by this closed surface is Qenclosed = (7dL)p. Using Gaussâ&#x20AC;&#x2122;s law,

In the last step, we have taken into account the radial direction of the electric field pointing outward. The potential difference between the surface and the axis is

30.84. Visualize: The distances from the point to the two charges are r+ and r-. These can be expressed in terms of the coordinates x and y by using the Pythagorean theorem. Y

Solve: (a) Potential is a scalar, so the potential at (x, y ) is the sum of the potentials due to each charge. This is 4 v = v+ + v- = -

4~&~r+

1 -

Potential and Field

30-25

(b) The binomial approximation is (1 + u)" = 1 + nu if u << 1. To use this, we'll need to rearrange the square roots of V to be in this form. The first term is

Jm~ + 1

= Jx2

+ y 2 - sy

= +[1+ X2

214

Jzj7

-sy + s 2 / 4 x?+y*

-1/2

i l +

-sy + s*/4 x2 + y *

1-q+s*/4 2+y2

1 =

Jm[1-5

The approximation for the second term yields the same expression except the term that is linear in y is negative. Substituting into the expression for the potential in part (a), V=

PY 4n&E,(x2 +y 2 y 2

This was a lot of algebra, but the resulting expression for Vis much simpler than the exact expression of part (a). (c) The electric field components are

(a) On-axis means on the axis of the dipole, or at x = 0 m. In this case, E, = 0 V/m and the electric field is in exact agreement with Eq. 26.1 1:

(e) The bisecting axis is the x-axis, so y = 0 m. In this case, E, = 0 V/m and the electric field is in exact agreement with Eq.26.12:

30.85. Visualize:

Solve: (a) The E-versus-r graph is shown above. (b) From Equation 30.3, the potential difference between the center (VJ and a point at radius r is

We can find V, from the reference point for the potential. At r = R the potential matches the known outside potential, so

30-26

Chapter 30

Using the expression in the AV equation, VR - v, = --- Q R2 4m0R3 2

* v, = --

Using this result for V,, the potential at radius r inside the cylinder is

(d) A V-versus-r graph is shown in the figure.

30.86. Visualize: For infinitely long cylinders, the electric field is radial from the inner cylinder to the outer cvlinder.

I 1 11-1 I I E

Solve: (a) The electric field in the space between the cylinders is the electric field of the charged inner cylinder. (Gaussâ&#x20AC;&#x2122;s law tells us that the outer cylinder does not contribute to the electric field between the cylinders.) From Chapter 16. we know that the field of a cylinder with linear charge density A is

The potential difference between the two cylinders is

The minus sign tells us that the potential decreases upon moving outward, but we only need the absolute value of AV. Capacitance is defined as C = Q/AV. Consider a segment of the cylinders of length L. The charge on the inner cylinder is Q = L. Consequently, the capacitance of this segment is

c=e= AV

2L .

(A/Zm,)In( R? / R, )

2z&, L In( R2 / R, )

The capacitance per meter is found by dividing out the L: capacitance per meter =

-= L

2z.5, In(R21Rl)

(b)Using the above expression, the capacitance per meter of cable is