Solutions to Mid-Term Exam for GP II,
MATH,
Spring 2012.
1. 25%
Two 1.20 m non-conducting wires meet at a right angle.
One segment carries +2.50 C of charge uniformly
distributed over its length and the other 2.50 C, likewise
distributed (see figure). Find the magnitude and
direction of the electric field at point P, which is 1.20 m
from each wire.
Solution:
Magnitude of field components on the line perpendicular to a line charge of
length L and going through its end point (x=0) is
L


Q1
1
Q L
x dx
Q
1
E//  k 
k  
 k 


3/2
L 0  x2  a2 
L a
L
L2  a 2 
x 2  a 2 0
k
Q 2 1
L2
2
if a  L
L

Q
Q L
a dx
aQ
x
k
E  k 
k


3/2
L 0  x2  a2 
L  a 2 x 2  a 2 0
a L2  a 2
k
Q
2 L2
if a  L
where a is the distance from the line and Q > 0 is assumed.
In the usual x-y coordinates, the field at P is
  E  E// 
E   1,1 , where
kQ  2  2  9  109  2.50  2  2 



 V / m
L2 
1.202
2 
2


 6.47  103 V / m
Magnitude of field is E  2   9.15  103V / m .
Direction is along the unit vector
e
1
1,1 .
2
2. 25%
A very long, solid insulating cylinder with radius R has a
cylindrical hole of radius a bored along its entire length.
The axis of the hole is a distance b from the axis of the
cylinder, where a  b  R (see figure). The solid part of
the cylinder has a uniform charge density . Find the
electric field E inside the hole.
Solution:
Let the axis of the cylinder be the z-axis. The field at a point r inside the cylinder is
in the x-y plane and in the radial direction. Its magnitude is given by the Gauss law
as


2 rE   r 2
r
 E
0
2 0
Hence, E 

r , where r is the position vector in the x-y plane.
2 0
If the cylinder axis goes through point b in the x-y plane, the field at r is simply

E
r  b  .
2 0
The field in the hole is just the superposition of the field of a solid cylinder of radius R
and charge density  on that of a solid cylinder of radius a , charge density  , and
at the position of the hole. Hence,



E
r
r  b  b
2 0
2 0
2 0
where b is the vector going from the axis of the cylinder to the axis of the hole.
3.
25%
Two plastic spheres, each carrying charge uniformly distributed throughout its
interior, are initially placed in contact and then released. One sphere is 60.0 cm
in diameter, has mass 50.0 g and contains 10.0 C of charge. The other is 30.0
cm in diameter, has mass 150.0 g and contains 40.0 C of charge. Find the
maximum acceleration and speed achieved by each sphere (relative to the fixed
point of their initial location in space), assuming no other forces are acting on
them.
Solution:
Since the spheres are outside of each other, their interactions can be calculated
by replacing them with point charges at their respective centers. Maximum
force occurs when they are closest:
10  106  40  106 
q1q2
9 
 17.8 N
F  k 2  9  10
2
r
 0.3  0.15
Max acceleration for charge
10.0 C :
a1 
F
 356 m / s 2
0.05
Max acceleration for charge
40.0 C :
a2 
F
 119 m / s 2
0.15
Maxima of speeds are obtained from
1. Energy conservation:
2. Momentum conservation:

U
1
1
m1v12  m2v22
2
2
m1v1  m2v2  0
1
m12  2
U   m1 
 v1
2
m2 
v1 
2Um2
m1  m1  m2 
10  106  40  106 
q1q2
9 
8J
U k
 9  10
r
0.3  0.15
Max acceleration for charge 10.0 C :
v1 
2  8.  0.15
 15.5 m / s
0.05  0.05  0.15
Max acceleration for charge 40.0 C :
m v 0.05  15.5
v2  1 1 
 5.17 m / s
m2
0.15
4. 25%
One plate of a parallel plate capacitor is connected
to the gold leaves of a Leyden jar. The other plate
of the capacitor and the metal foil on the outside of
the jar are both grounded (see figure). A charge Q
is then introduced to the capacitor plate that is not
grounded. The gold leaves are observed to spread
apart by an angle . If the capacitor plates are
now moved further apart, how does  change? Derive your result in detail.
Solution:
Treating the Leyden jar as a capacitor of capacitance CL , the system becomes 2
capacitors in parallel connection. Let q and qL be the final charges on the
capacitor and Leyden jar, respectively. Thus
q  CV
qL  CLV
q  qL  Q

q CL Q  qL 

C
C
CL
qL 
Q
C  CL
qL  CL

Pulling the capacitor plates further apart lowers C so that qL becomes larger,
which in turn increases .
Table of Integrals
Let r  x 2  a 2 ,

dx
 ln  x  r 
r

xdx
r
r
x
a2 r

xdx
1

3
r
r
dx
r
3
