Physics 21
Fall, 2009
Solution to HW6
2340
Two large, parallel conducting plates carrying op
posite charges of equal magnitude are separated by
d
=
2
.
20 cm.
(a) If the surface charge density for each plate
has magnitude 47.0 nC/m
2
, what is the magnitude of
E
in
the region betwixt the two plates? (b) What is the poten
tial difference between the two plates? (c) If the separation
between the plates is doubled while the surface charge den
sity is kept constant at the value in part (a), what happens
to the magnitude of the electric field and to the potential
difference?
(a) Two parallel plates with opposite surface charge form a
capacitor. We have discussed the field
E
between capacitor
plates in terms of the field
σ/
2
0
of a single plate. From the
equation sheet, the field between the plates is
E
=
σ
0
=
47
.
0
×
10

9
Cm

2
8
.
85
×
10

12
C
2
N

1
m

2
= 5310
N
C
(b) To find the potential difference, integrate along the sim
plest path between the plates: a straight line from the neg
ative to the positive plate and perpendicular to both.
For
this path, the potential will increase since the direction of
integration is opposite to the field.
V
=
−
E
·
dl
=
Ed
=
5310
N
C
(
.
022 m) = 117 V
(c) The electric field
E
doesn’t depend on the distance be
tween plates; it only depends on
σ
.
For this problem,
σ
stays the same as
d
is increased. Therefore the electric field
remains constant, and the potential difference
Ed
increases.
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 Spring '07
 Hickman
 Charge, Potential difference, Electric charge, Felectric

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