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Question set E E(1) (a) Write down the Coloumb’s law of force.

Coulomb’s law of force claims, that interaction force between two point charges in vacuum directed along the line, connecting those particles, proportional to their chagres q1 and q2 and inversely proportional 2 . This is attraction force, if charges have different signs, and to the squared distance between them r1,2 repulsion, if charges have similar sign. Mathematically it is: F = ke

q1 q2 , 2 r1,2

(E.1.1)

1 ≈ 9 × 109 N· m2 /C2 is so called Coulomb constant. where ke = 4πε 0 (b) Two electrons of charge −1.6 × 10−19 C are 0.001 m apart. What is the force on each?

Accordingly to formula (E.1.1), the force will be equal

F = 9 × 109

(−1.6 × 10−19 )2 = 2.3 × 10−23 N. (1 × 10−3 )2

(c) Write down Gauss’ theorem in Electrostatics.

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Mathematically this can be written as I Q (E.1.2) E · dA = , ε0 S where E is the electric field, dA is a vector representing an infinitesimal element of area, and represents the dot product. (d) By using a suitable Gaussian surface, find the field E at a distance d from a very large charged plane charged to uniform charge density σ C/m2 .

To find the E-field will build a cylinder with bases symmetrical to the plane as shown on a figure.

If area of base is S, electrical flux through one base equals ES, through both – 2ES. The flux through later surface equals 0, as E and n are perpendicular. Thus, the flux equal Φ = 2ES From other side, according to the Gauss’ theorem, it equals: Φ=

q σS = ε0 ε0

2 After equating those two expressions, will get: E=

σ 2ε0

As we can see, electric field doesn’t depend on the distance to the plane. The direction is out from the plane. (e) What is the magnitude and direction of force on a small particle charged to −3 × 10−5 C at a distance 6 mm from a very large plane charged to a uniform surface charge density σ = 5 C/m2 ?

Accordingly to the expression, recieved in previous point, force equals F = qE =

−3 × 10−5 × 5 qσ = = 8.5 × 106 N −12 2ε0 2 × 8.85 × 10

As the plane and partical have different charge sign, that would be an attraction force. (f ) A metallic sphere of radius R has a unifrom surface charge density σ C/m2 . What is the electric field in distance r from the centre of the sphere for: r > R and r < R. As the sphere has the charge only on its surface, the field inside it equals 0. (Any surface radius less then radius of sphere will contain 0 charge.)

From Gauss’ theorem Q = ES ε0 2 σ R 4πR2 σ = E= 4πr2 ε0 ε0 r

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E(2) The figure shows a cross section of a parallel plate capacitor. The plates are of area A and the distance between them is a+b as shown. The space 0 < y < b is filled with dielectric of relative permittivity ε2 and the space b < y < a + b is filled with dielectirc of relative permittivity ε1 .

(a) Draw on a suitably labelled sketch the E field between the plates.

(b) Draw on a suitably labelled sketch the D field between the plates.

(c) Using Gauss’ Flux Theorem, or some other suitable method, derive an expression for the E field in the two regions: 0 < y < b, b < y < a + b To find the E-field (in region 0 < y < b) will build a cylinder with bases symmetrical to upper plane as shown on a figure.

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If area of base is S, electrical flux through one base equals ES, through both – 2ES. The flux through lateral surface equals 0, as E and n are perpendicular. Thus, the flux equal Φ = 2ES From other side, according to the Gauss’ theorem, it equals: Φ=

q AQf = ε0 ε0 ε2

After equating those two expressions, will get: AQf 2ε0 ε2

Eup =

From the same considerations we will get the field created by the lower plane: Edown = −

AQf 2ε0 ε2

E1 = Eup − Edown =

AQf ε0 ε2

Similary, E2 =

AQf ε0 ε1

(d) Calculate the capacitance of the device.

C=

QF U

Za+b Zb Za+b AQf a b U= Edy = E1 dy + E2 dy = E1 a + E2 b = ( + ) ε0 ε2 ε1 0

0

b

C=

ε0 a b ( + )−1 A ε2 ε1

(e) Calculate the capacitance if A = 1200 mm2 , a = 50 mm, b = 55 mm, ε1 = 1.6 and ε2 = 2.6.

C=

8.85 × 10−12 0.05 0.055 −1 ( + ) = 140 nF. 1.2 × −3 2.6 1.6

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E(5) The figure shows a small steel electromagnet. Two coils each have 200 turns in series and produce magnetic flux in the same direction. Dimensions of magnet are as follows: l1 = 250 mm, l2 = 65 mm, lg = 3 mm, area A1 = 1500 mm2 , and area A2 = 3700 mm2 .

(a) Sketch an equivalent magnetic circuit for finding magnetic flux in magnet.

(b) Assume a linear magnetistion model for the steel with relative permeability of 800; find the magnetic flux density of the air gap for current of 10 A in the coils.

From the Ampere’s circuital law 2N I = H1 l1 + H2 l2 + Hg lg From the Gauss’ theoreme for magnetic field: I

~ s=0 Bd~

S

Φg = Φ2 = 2Φ1 2B1 A1 = Bg A2 = B2 = A2

6 B2 = Bg = B B1 = B

A1 2A2

−1 l2 l1 A1 B = 2N Iµo lg + + ≈ 1.6T. µ µ 2A2 (c) Now assume a nonlinear magnetic property for the steel magnet instead of linear model. The relative magnetisation curve for the magnet steel is given by the expression: B=

0.95H 1445 + H

Find the current in the coils for an air gap flux density of 0.4 T.

B lg 1445 289 I= + + ≈ 6.14A. 2N µo 0.95 − B 0.95 − B/5 (d) State what assumptions you have made in parts (5.b) and (5.c) to be able to use a magnetic circuit model.

Used Ampere’s law and Gauss’ theorem.