SOLUTIONS – HW3:
Q2214) The total charge that is enclosed in case (a) is +1.0
m
C – 2.0
m
C = 1
m
C, in
case (b) it is zero (as object 3 is neutral). Thus, using Gauss’s law:
0
6
0
10
0
.
1
ε
ε
C
Q
encl
E
−
×
−
=
=
Φ
in case (a) and zero in case (b).
Q236) If V = 0 at a point in space, then the electric field does not have to be zero,
even if
V
E
−∇
=
r
. Remember, you take the derivative first, then you evaluate it at
that point. Example: V(x) = x is zero at x = 0, but E(x) = dV(x)/dx = 1 is not zero
at x = 0.
If E = 0 at a point in space, then the potential V does not have to be zero at that
point. Example: E(x) = 1x is zero at x = 1. V = V0 + x – ½ x
2
= V0 + ½ at x = 1
does not have to be zero as we can chose V0 to fulfill our boundary condition.
Q2318) If the electric field is uniform, the integral will simplify to E L, where L is the
distance between two points, and the potential difference between these two points
is E L. If we choose V = 0 at L = 0, then the potential increases linearly with
increasing distance L.
If V is uniform in a region of space, V = V0 = constant and
V
E
−∇
=
r
=0.
Q246) The electrons flow from the negative terminal of the battery to one plate of
the capacitor. At the same time, electrons flow from the second plate to the positive
terminal of the battery. This will still be true of the two conductors have different
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 Spring '08
 Jordan
 Electron, Charge, Electric charge

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