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(phys121)[2009](f)midterm~ma_yxf^_10541.pdf
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Solution 1:
(a) E = 0 , V = 04 1 R Q

1
. ..
.

.

22 2
. Q
2
.

1 Q

.

Q

1

z

.

(b)

U (z)

..

=

=
4
R

2
4

22
R

2
R

2R

(/ R)

+

+

z

z

0
0 0
2 22 22
. Q Qz Qz
=+ = + cons tan t
4 R 8 R38 R3
00 0
Compare it with the potential energy of a harmonic oscillator
12
U (z) = Kz + cons tan t
2
We show that the motion of charge with small displacements is indeed a simple harmonic
Q2 oscillation with a spring constant of K =
4 0 R3
Alternative solution for (b) starting from the electric field:
Qz
E()z = 22
2
40(R + z )3
Thus a negative charge CQ at point on the z axis for z much smaller than R is acted upon by a force:
. Q2 zz<<R . Q2 z
F = ..
=.Kz
2 23/2 3
4 0(R + z )4 0 R As z << R the force is in a good approximation proportional to z and CQ will execute simple harmonic motion.
Solution 2:
(a) The capacitance of the conducting soap bulb is C = 4 0 r , and the stored electric energy is
Q2 Q2
U == .
2C 8 0 r The electric force can be obtained by
F dA dA dU dA Q2
dF = dA = F =. =
222 22
4r 4r 4r dr 4r 8 0 r
QdA Q Q 1

== dq =
Edq
22 2
4r 8 0 r 8 0 r 2

Alternative solution for (a):
The surface charge density is given by
Q
=
4 r 2
We consider a point P inside the sphere close to an area element dA. The charge dq on this area element will produce at the point P an electric field which is approximately that due to a uniformly charged infinite plate, namely,

E =. n
1 p 20
Where n is a unit vector normal to dA in the outward direction.
The electric field is zero inside a conducting sphere. Hence, if we take E2 p as the the electric field

at P due to all the charges on the spherical surface except from the element ds, we must have
E = E + E = 0

p 1 p 2 p
Q
Therefore E = n = n
2 p 2
208 0r
As P is close to dA, E2 p may be considered as the field strength at ds due to the charges of the
spherical surface. Hence, the force acting on d s is 1
dF = dqE 2P = Edq n ,
2
Q
Where E = 2 is just the field strength on the spherical surface. 4 0r
(b) Charge is conserved Q = CiVi
2
Q 1212 2 .9
initial energy U == CV =
4rV = 2 rV = 5.56 10 J
ii 0 i 0 i
2Ci 22
Q2 Q2 Ci ri .8
Final energy Uf == = Ui
= 5.56 10 J
2C 2CC r
fif f
Explanation: As the soap bulb collapses, an external force must do a positive work to overcome the radial electric force and this work is converted into the electric energy. Or equivalently, as the soap bulb collapses, the radial electric force does negative work to increase the electric potential energy.
Solution 3:
(a)
E=0 for r < a . Otherwise, free electrons would move inside the conductor.

(b)
Lets assume in the left and right sides, E (r) =/ r , E (r) = / r


11 22 bb 22
Because a conductor is in equ