Download Solutions to Mid-Term Exam for GP II, HOE, Spring 2011. 25% In the

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
10%
A non-conducting square plate 50 cm on a side carries a uniform surface charge
density. The electric field strength 1 cm from the plate, not near the edges, is
30 kN/C. What is the approximate field strength 20 m from the plate?
Solution
Far from the edges & close to the plate, we have
  2 0 E  60  103  0 N/C
Far away, the plate looks like a point charge:
E

so that
2 0
60  103  0   0.5
 A
 3 N/C
E

2
4 0 r 2
4 0  20 
2
4.
15%
A 2.0-cm-radius metal sphere carries 75 nC and is surrounded by a concentric
spherical conducting shell of radius 10 cm carrying 75 nC.
(a) Find the potential difference between the shell and the sphere.
(b) How would your answer change if the shell charge were changed to +150
nC?
Solution
(a) For 2.0 cm  r  10 cm ,
4 r 2 E 
75  109
Vshell  Vsphere
(b)
0

E
75  109
4 0 r 2
1
75  109
  k
dr  9  109  75  109
2
0.02
r
r
0.1
0.1
 27000 V
0.02
Unchanged.
5. 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.
So the field between them is everywhere
E
V
, where d is distance between the plates.
d
The charge densities on the
left & right side are however different:


E L  R
 0 0
Hence, the charge on the upper plate is
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