ELECTRIC AND MAGNETIC FIELDS
ANSWERS TO ASSIGNMENT 5
Q1:
Ionosphere
30
(a) Energy density of the electric field is
u
1
0E2.
2
E
120 km
Ground
10 for formula
In this case, E is constant: E = 100 V m-1 
u = 4.43 x 10-8 J m-3.
5 for result
(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)
15 = 10 for method
and 5 for result
 Utot = (4.43 x 10-8)[4(6.4 x 106)2](1.2 x 105) = 2.73 x 1012 J.
Slightly more accurate calculation:
Vol. of atmosphere =
4
 (R E  120km) 3  (R E ) 3 = 6.29 x 1019 m3
3
 Utot = 2.79 x
1012 J.
Q2:
1
The energy of a system of point charges is
35
U tot
2
a
Q
1 n
  Q i Vi .
2 i 1
Q
2 a
By symmetry,
a
8 for good
diagram
 All four charges are equal to Q, and they
all
have the same energy. Therefore,
Q
Utot = (1/2)(4)QV1 = 2QV1
Q
4
where V1 is the potential at charge 1 (for
instance) due to all of the other charges.
10 for getting
to here or equivalent
 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/(4oa)
3
V13 = Q/(4oa2)
10 for knowing
what VI means
and expressing
it as three
contributions
So U tot

 Q
Q
Q 
Q2 
2 
Q2
 2Q 


4 2

2  2 

2  40 a
 40 a 40 a 2 40 a  40 a 

7 for
final
answer
Q3: (i)
By spherical symmetry, the field lines are
radial.
30
E = 0 inside the solid conducting sphere.
E
Field lines originate on positive charges on the
surface of the inner conductor, and terminate
R2
on
negative charges on the inner surface of the
R1
shell.
(ii)
given
The energy density of the electric field is
by
u = ½0E2
J m-3
3
The field in the region between the conductors is the same as for a point charge,
because the charge distribution is spherically symmetric:
6 for good diagram
(note – question doesn’t
Q
3
E
ask for explanation
4 0 r 2
1
Q2
3
The energy density is therefore u   0 E 2 
2
32  2  0 2 r 4
(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
Shell thickness
dr
d(Volume) = 4
r2dr
r
The energy contained in the shell is
.
8 for good
explanation
and
getting to
the integral
+
Q2
dU  u( r )d( Volume ) 
4 r 2 dr
2
4
32   0 r
The total energy is obtained by integrating
this over the region in which E exists - i.e.,
Therefore
5 for
integrating
to get
to the
answer
between R1 and R2.
b
Q 2 -2
Q2
r dr 
80
80
a
U tot  
(iv) If b >> a then U tot 
1 1 
 a  b   U tot 


Q2
8 0
b  a
 ab 
Q2
 formula derived in the lectures for total
80 a
electric energy of an isolated conducting sphere of radius a.
2
Q4:
5
Y
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
Y
E
L
L
2
A
1
B
3
4
0
L
.If E is conservative, then

X
X
E  d L  0 (or
0
L
P

E .d L is the same for paths A and B).
0
We can integrate E  dL around the square loop. Splitting the integral into four sections,
we can see from the filed pattern that:
Sections 1 and 4 : E and dL are perpendicular
Section 2
: The x component of E is parallel to dL
positive
Section 3
: The y component of E is anti-parallel to dL
negative


E  dL 
2
E  dL 
1
E  dL 
1



E  dL 




E  dL 
2
(ay ˆi  bx ˆj)  dy 
(ay ˆi  bx ˆj)  dx 
E  dL 
3

3


E  dL is
E  dL
( bx ˆj)  dy  0

E  dL = 0
E  dL is
4
(ayˆi )  dx 
as x = 0

(aL ˆi )  dx =

The length of the path is L, so
E  dL 




2


aLdx = aL dx as y = L and ˆi  dx = dx
E  d L  aL2



(ay ˆi  bx ˆj)  dy   ( bx ˆj).dy   (bL ˆj).dy = - bLdy = -bL dy
The length of this path is L, so

2
E  dL   bL2
as x = L and ˆj  dy = dy
Difficult – some
may get this as positive.
Don’t penalise harshly

E  dL 
4
Therefore

( ay ˆi  bx ˆj)  dx  -

( ay ˆi )  dx  0
as y = 0