Download FinalExamII_Solution

Solutions to Final Exam for General Physics II, Spring 2010.
1.
20%
A point charge q is at the center of a spherical shell carrying charge +2q.
That shell, in turn, is concentric with a larger shell carrying charge 3q/2.
Draw a cross section of this structure, and sketch the electric field lines using the
convention that eight lines correspond to a charge of magnitude q.
Solution
2. 20%
An infinitely long rod of radius R carries a uniform volume charge density .
Find an expression for the electrostatic energy per unit length contained within
the rod.
Solution
E at a point r from the axis is given by applying Gauss’ law to a cylindrical
volume of cross section  r 2 and length L :
E 2  r L 
  r 2  L
0
E

The energy density at that point is
r
2 0
1
 2 r2
uE   0 E 2 
2
8 0
The energy within a length L of the rod is
2
U  2 L  uE r d r  2 L
8 0
0
R
Energy per unit length is u 
R

r3 d r   L
0
U
 2 R4

L
16 0
 2 R4
16 0
3.
20%
In the circuit shown below, it makes no difference whether the switch is open or
closed. What is 3 in terms of the other quantities shown?
Solution
Switch open, outer loop:
I R1  E1  E2  I R2  0
Switch closed, left-hand loop:
I1R1  E1  E3  0
I  I1
4.

E3  I R1  E1 
R E  R2 E1
E1  E2
R1  E1  1 2
R1  R2
R1  R2
20%
The structure shown in the figure below is made from conducting rods.
The upper horizontal rod is free to slide vertically on the uprights, while
maintaining electrical contact with them.
The upper rod has mass m and length L.
A battery connected across the insulating gap at the bottom of the left-hand
upright drives a current I through the structure.
At what height h will the upper wire be in equilibrium?
Express your answer in terms of the given quantities.
Solution
Currents flow in opposite directions in the horizontal bars so that they experience
a repulsive force
F
0 I 2 L
2 h
In equilibrium F  mg

h
0 I 2 L
2 m g
5. 20%
Resistor R2 in the circuit below is to limit the emf that develops when the switch
is opened. What should its value be so that the inductor emf does not exceed
100 V ?
Solution
When the switch is thrown open, the emf across the inductor is
EL  I R2
where I is the current flowing through the inductor just before the switch is
opened. Hence,
max EL  I max R2
where
Imax
is the maximum possible current through the inductor.
Imax occurs when the switch is closed for a long time so that the inductor is
equivalent to a wire.
45 V
1
E

 A
I max 
R1 180  4
Hence,
R2 
100 V
 400  .
1
A
4