P.41
The upper and lower conducting plates of a large parallelplate capacitor are separated by a distance
d
and maintained at potentials
V
0
and 0, respectively. A dielectric slab of dielectric constant 6
.
0 and
uniform thickness 0
.
8
d
is placed over the lower plate. Assuming negligible fringing effect, determine
(a) the potential and electric field distribution in the dielectric slab,
(b) the potential and electric field distribution in the air space between the dielectric slab and
the upper plate,
(c) the surface charge densities on the upper and lower plates.
(d) Compare the results in part (b) with those without the dielectric slab.
d
0.8d
V=V0
V=0
Solution: Neglecting fringing, the potential in each region is the solution Laplace’s equation in one
dimension:
∇
2
V
=
d
2
V
dx
2
= 0
Hence, we have
V
(
x
)
=
braceleftBigg
c
1
x
+
c
2
, 0
≤
x
≤
0
.
8
d
c
3
x
+
c
4
, 0
.
8
d < x
≤
d
The resulting electric field in each region is given by
E
=
braceleftBigg
−
ˆ
a
x
c
1
, 0
≤
x
≤
0
.
8
d
−
ˆ
a
x
c
3
, 0
.
8
d < x
≤
d
and the electric flux density by
D
=
braceleftBigg
−
ˆ
a
x
ǫ
r
ǫ
0
c
1
, 0
≤
x
≤
0
.
8
d
−
ˆ
a
x
ǫ
0
c
3
, 0
.
8
d < x
≤
d
The boundary conditions to be applied to obtain
c
1
,
c
2
,
c
3
, and
c
4
are
(1)
V
(0) = 0
(2)
V
(
d
) =
V
0
(3)
V
(0
.
8
d

) =
V
(0
.
8
d
+
)
(4)
D
x
(0
.
8
d

) =
D
x
(0
.
8
d
+
)
2
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The resulting system of equations can be written as
0
1
0
0
0
0
d
1
0
.
8
d
1
−
0
.
8
d
−
1
ǫ
r
0
−
1
0
c
1
c
2
c
3
c
4
=
0
V
0
0
0
Solving the linear system, we obtain
c
1
=
V
0
d
(0
.
8 + 0
.
2
ǫ
r
)
c
2
=
0
c
3
=
ǫ
r
V
0
d
(0
.
8 + 0
.
2
ǫ
r
)
c
4
=
0
.
8 (
ǫ
r
−
1)
V
0
(0
.
8 + 0
.
2
ǫ
r
)
To solve parts (a), (b) and (c) with the dielectric present, we set
ǫ
r
= 6 in the above. To solve part
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 Spring '07
 ayanar
 Electrostatics, Electromagnet, Electric charge, Fundamental physics concepts, dielectric slab

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