OUTLINE OF SOLUTIONS
PHYS-102 SUMMER-08 QUIZ-II Time Limit: 50 min.
Notes:1. For problems 2 and 3, your solutions must have adequate details to get full credit.
F = qE ; 12
2
kQ Q
Fr
= ; 1/4ฯฮต0 = 92
910kxNmC
2
โ
=, U = kQ1Q2/r
V = kQ/r ; 1/Cser = 1/C1+ 1/C2+1/C3+โฆ. ; C = Q/V
Cpar. = C1 + C2 +C3+โฆ.; U = QV/2 ; -
ฮ
Wfield =
ฮ
U =
ฮ
Wext ; Ex = - dV/dx
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NOTE: For Prob.1, below, please circle the answer of your choice for each part.
1a.( 3pts.)Consider two conducting metallic spheres S1 and S2. The radius of S1 is half that of
S2. Each sphere is given an excess positive charge Q. If the two spheres now are brought in
electrical contact:
[a] charge will flow from S1 to S2.( since V = kQ/r, SI will be at a higher potential than S2)
[b] charge will flow from S2 to S1.
[c] no charge flow will take place.
[d] the charge on each sphere will double.
1b(3pts.) When the potential difference between the plates of a capacitor is doubled, the
magnitude of the electric energy stored in the capacitor:
[a] is doubled [b] is halved [c] remains the same [d] quadrupled (U = QV/2= CV2/2)
1c (3pts) The electric potential, V(x) in a certain region of space is given by the
expression V(x) = 3.0 + 4.0x, where x is measured in meters and V(x) in Volts. The value
of the E-field at x = 2.0 m will be (in units of V/m):
[a] 11.0 [b] โ 5.0 [c] โ 4.0 (Ex = - dV/dx) [d] 0.0
1d(3pts) Two charges are placed along the x-axis as
shown. Now, suppose you move the positive charge
( while keeping the negative charge at the origin) to
a point 2X away from the origin. As a result of this
the electric potential energy of the two charges wou
ld:
-Q +Q
X
[a] increase (you will have to do work to move the charges and hence increase the
potential energy or you can simply use U = kQ1Q2/r to see that the potential energy
increases when you double the separation.)[b] decrease [c] remain the same [d] be
reduced to zero
