P26_047.PDF

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Chapter 26 - 26.47
q , we find an expression for
the electric field in each region, in terms of q , then use the result to find an expression for the potential
difference V between the plates. The capacitance is
−
C =
q
V
.
The electric field in the dielectric is E d = q/κε 0 A ,where κ is the dielectric constant and A is the plate
area. Outside the dielectric (but still between the capacitor plates) the field is E = q/ε 0 A . The field is
uniform in each region so the potential difference across the plates is
V = E d b + E ( d
−
b )=
κε 0 A + q ( d
qb
−
b )
=
q
ε 0 A
b + κ ( d
−
b )
.
ε 0 A
κ
The capacitance is
q
V
κ ( d − b )+ b = κε 0 A
κε 0 A
κd − b ( κ − 1) .
The result does not depend on where the dielectric is located between the plates; it might be touching
one plate or it might have a vacuum gap on each side.
C =
=
For the capacitor of Sample Problem 26-8, κ =2 . 61, A = 115 cm 2 = 115
×
10 − 4 m 2 , d =1 . 24 cm =
1 . 24
×
10 − 2 m, and b =0 . 78 cm = 0 . 78
×
10 − 2 m, so
C =
2 . 61(8 . 85
×
10 − 12 F / m)(115
×
10 − 4 m 2 )
2 . 61(1 . 24
×
10 − 2 m)
−
(0 . 780
×
10 − 2 m)(2 . 61
−
1)
=1 . 34
×
10 − 11 F=13 . 4pF
in agreement with the result found in the sample problem. If b =0and κ = 1, then the expression
derived above yields C = ε 0 A/d , the correct expression for a parallel-plate capacitor with no dielectric.
If b = d , then the derived expression yields C = κε 0 A/d , the correct expression for a parallel-plate
capacitor completely filled with a dielectric.
47. Assuming the charge on one plate is + q and the charge on the other plate is
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