Solutions to
1.
Mid-Term Exam for GP II, HOE, Spring 2011.
25%
In the figure on the left, point i represents the
initial state of an ideal gas at temperature T.
Calculate and rank the entropy changes in going
from i to the points a, b, c, and d.
Solution
V  0  W  0 .
Q  U  CV T
(a)
Q
S   CV
T

T
dT
T  T 3
T
 CV ln
 nR
T
T
2
T
Q  CP T
(b)
Q
S   CP
T
T T

T
dT
T  T 5
T
 CP ln
 nR
T
T
2
T
3
T
S   n R
2
T
(c)
(d)
T T
5
T
S   n R
2
T
Ranking is (b)  (a )  ( c)  ( d )
2. 25%
A thin rod of length L has a line charge density given by
4 x2
2 x
sin
2
L
L
where 0 is a constant and x  0 is at the center of the rod. Find the electric
  0
field and potential at x  0 , assuming the zero potential to be at infinity .
Hint:
Solution
Consider contributions from the left and right sides separately.
L /2
 0 
 
E  k   2 dx   2 dx  i
x
0
  L /2 x

0
L /2
4 
2 x
2 x 
 k 0 2   sin
dx   sin
dx  i
L   L /2
L
L
0

Since
0

 L /2

2 x
2 x
dx    sin
dx
L
L
0
L /2
sin
8
E  k 0 2
L
L /2
 sin
0
2 x
8 L
8
dx i  k 0 2
i
 cos   1 i  k 0
L
L 2
L
L /2
 0 
 
k  
dx  
dx 
x
0
  L /2 x

0
L /2
4 
2 x
2 x 
 k 0 2   x sin
dx   x sin
dx 
L   L /2
L
L
0

Since
0


 L /2

3.
2 x
2 x
dx    x sin
dx
L
L
0
L /2
x sin
  x  0  0
25%
A solid sphere 10 cm in radius carries a uniform 40 C charge distribution
throughout its volume. It is surrounded by a concentric shell 20 cm in radius,
also uniformly charged with 40 C. Find the electric field at
(a) 5.0 cm,
(b) 15 cm, and
(c) 30 cm
from the center.
Solution
(a)
A point at 5.0 cm is inside the solid sphere. Applying Gauss’ law gives
4 r 2 E 
 4 r 3 Q r 3

0 3
 0 R3
E  9  109 
40  106  0.05
 0.1
3

E
Qr
Qr
k 3
3
4 0 R
R
N/C  18  106 N/C
(b)
A point at 15.0 cm is outside the solid sphere but inside the shell.
Ek
Q
40  106
9
 1.6  107 N/C

9

10

2
2
r
 0.15
(c)
A point at 30.0 cm is outside both the solid sphere and the shell.
 40  40  10
Q
E  k 2  9 109 
2
r
 0.30 
6
 8  106 N/C
4. 25%
An air-insulated parallel-plate capacitor of
capacitance C0 is charged to voltage V0 and then
disconnected from the charging battery. A
slab of material with dielectric constant ,
whose thickness is essentially equal to the
capacitor spacing, is then inserted a distance X
into the capacitor (see figure). Determine
(a) the new capacitance, and
(b) the stored energy.
Neglect all fringe effects.
Solution
(a)
Each plate is an equipotential surface.
E
So the field between them is everywhere
V
, where d is distance between the plates.
d
left & right side are however different:


E L  R
 0 0
Hence, the charge on the upper plate is
The charge densities on the
q    L X   R  L  X   W   X  L  X   0 E W
The new capacitance is therefore
C
q
W
  X  L  X   0
 C0
V
d
where C0   0
LW
d
X

 1    1 
L

is the original capacitance.
(b)
Since the battery is disconnected, q  q0 . The stored energy is
q02

U
2C
where
U0 
q02
X

2C0  1    1 
L


U0
1    1
q02
is the original stored energy.
2C0
X
L