ELECTRIC AND MAGNETIC FIELDS ANSWERS TO ASSIGNMENT 5

Q1: (a) Energy density of the electric field is

u E=12 0

2ε .

In this case, E is constant: E = 100 V m-1 ⇒ u = 4.43 x 10-8 J m-3. (b) The total electric energy of the atmosphere is Utot = (Energy density)(Volume of the atmosphere) ⇒ Utot = (Energy density)(Area of Earth's surface)(Height of ionosphere) ⇒ Utot = (4.43 x 10-8)[4π(6.4 x 106)2](1.2 x 105) = 2.73 x 1012 J. Slightly more accurate calculation:

Vol. of atmosphere = 43

120 3 3π ( ) ( )R km RE E+ − = 6.29 x 1019 m3 ⇒ Utot = 2.79 x

1012 J. Q2: The energy of a system of point charges is

U Q Vtot i ii

n =

=∑

12 1

.

By symmetry, • All four charges are equal to Q, and they

all have the same energy. Therefore, Utot = (1/2)(4)QV1 = 2QV1 where V1 is the potential at charge 1 (for

instance) due to all of the other charges. • V1 has three contributions, from charges 2, 3, and 4: V1 = V12 + V13 + V14 where V12 = potential at position 1 due to the charge at position 2, etc. • V12 = V14 = Q/(4πεoa) V13 = Q/(4πεoa√2)

Ionosphere

Ground

120 kmE

a

a

Q

Q

Q

1 2

34

Q

2 a

30

10 for formula

5 for result

15 = 10 for method and 5 for result

8 for good diagram

35

10 for getting to here or equivalent

10 for knowing what VI means and expressing it as three contributions

So [ ]24a4

Q2

222a4

Qa4

Q2a4

Qa4

QQ2U0

2

0

2

000tot +

πε=

++πε

=

πε+

πε+

πε=

7 for final answer

Q3: (i) By spherical symmetry, the field lines are radial. E = 0 inside the solid conducting sphere. Field lines originate on positive charges on the surface of the inner conductor, and terminate on negative charges on the inner surface of the shell. (ii) The energy density of the electric field is given by u = ½ε0E2 J m-3 The field in the region between the conductors is the same as for a point charge,

because the charge distribution is spherically symmetric:

E Qr

=4 0

2πε

The energy density is therefore u E Qr

= =12 320

22

20

2 4επ ε

(iii) The total energy is obtained by integrating this expression over the entire volume

between a and b. As the incremental volume element, choose a spherical shell of radius r, thickness dr.

Volume of shell is d(Volume) = 4πr2dr The energy contained in the shell is

dU u r d Volume Qr

r dr = =( ) ( )2

20

42

324

π επ

The total energy is obtained by integrating this over the region in which E exists - i.e., between R1 and R2. Therefore

R2

E

R1

drShell thickness

r

.

30

6 for good diagram (note – question doesn’t ask for explanation

3

3

3

8 for good explanation and getting to the integral + 5 for integrating to get to the answer

-

−

πε=

πε= ∫ b

1a1

8Qdrr

8QU

0

22

b

a 0

2

tot ⇒ U Q b aabtot =−

2

08πε

(iv) If b >> a then a8

QU0

2

tot πε= ≡ formula derived in the lectures for total

electric energy of an isolated conducting sphere of radius a.

2

Q4:

.If E is conservative, then ∫ =⋅ 0d LE (or ∫P

0

d L.E is the same for paths A and B).

We can integrate LE d⋅ around the square loop. Splitting the integral into four sections, we can see from the filed pattern that:

Sections 1 and 4 : E and Ld are perpendicular ⇒ LE d⋅ = 0 Section 2 : The x component of E is parallel to Ld ⇒ LE d⋅ is positive Section 3 : The y component of E is anti-parallel to Ld ⇒ LE d⋅ is negative

0xas0d)ˆbxd)ˆbxˆayd

ddddd

= yj( yji( LE

LE LE LE LE LE

1

4321

=⋅=⋅+=⋅

⋅+⋅+⋅+⋅=⋅

∫∫∫∫∫∫∫∫

dxdˆandLyasdxaLaLdxd)ˆaLd)ˆayd)ˆbxˆayd =xi = = = xi( xi( xji( LE

2

⋅⋅=⋅=⋅+=⋅ ∫∫∫∫∫∫

The length of the path is L, so 2

2

aLd LE =⋅∫

dydˆandLxasdybLbLdyd).ˆbL(d).ˆbxd)ˆbxˆayd

3

=yj = -=-=yjyj(yji( LE ⋅−=−=−⋅+=⋅ ∫∫∫∫∫∫

The length of this path is L, so 2

2

bLd −=⋅∫ LE Difficult – some may get this as positive. Don’t penalise harshly

5

L

L

0

Y

1 2

3

4 X

L

L

0

Y

A

B

X

E It is very useful to draw the electric field vector at various positions, as shown

~ 3 for having the right idea ~ 2 for all the details

0yas0d)ˆayd)ˆbxˆayd = xi(- xji( LE

4

=⋅=−⋅+=⋅ ∫∫∫

Therefore