Page 324 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 26] CAPACITANCE 309

The constant ε 0 is the permittivity of free space mentioned in Chapter 23; its value is
2
2
ε 0 = 8.85 × 10 −12 C /(N·m ) = 8.85 × 10 −12 F/m
The quantity K is the dielectric constant of the material between the capacitor plates; the greater K is, the more
effective the material is in diminishing an electric field. For free space, K = 1; for air, K = 1.0006; a typical
value for glass is K = 6; and for water, K = 80.

SOLVED PROBLEM 26.1
A 200-pF capacitor is connected to a 100-V battery. Find the charge on the capacitor’s plates.

Q = CV = (200 × 10 −12 F)(100 V) = 2 × 10 −8 C

SOLVED PROBLEM 26.2
A capacitor has a charge of 5 × 10 −4 C when the potential difference across its plates is 300 V. Find its
capacitance.

Q 5 × 10 −4 C
C = = = 1.67 × 10 −6 F = 1.67 µF
V 300 V
SOLVED PROBLEM 26.3

A parallel-plate capacitor has plates 5 cm square and 0.1 mm apart. Find its capacitance (a) in air and
(b) with mica of K = 6 between the plates.
(a) In air K is very nearly 1, and so

A −12 F (0.05 m) 2 −10
C = Kε 0 = (1) 8.85 × 10 −4 = 2.21 × 10 F = 221 pF
d m (10 m)
(b) With mica between the plates the capacitance will be K = 6 times greater, or
C = (16)(221 pF) = 1326 pF

SOLVED PROBLEM 26.4
A parallel-plate capacitor has a capacitance of 2 µF in air and 4.6 µF when it is immersed in benzene.
What is the dielectric constant of benzene?
Since C is proportional to K, in general

C 1 C 2
=
K 1 K 2
for the same capacitor. Here, with K 1 = K air = 1, the dielectric constant K 2 of benzene is

C 2 4.6 µF
K 2 = K 1 = (1) = 2.3
C 1 2 µF
SOLVED PROBLEM 26.5
A 10-µF capacitor with air between its plates is connected to a 50-V source and then disconnected.
(a) What is the charge on the capacitor and the potential difference across it? (b) The space between the
plates of the charged capacitor is filled with Teflon (K = 2.1). What is the charge on the capacitor and
the potential difference across it now?