Question

An air-filled capacitor is made from two flat parallel plates 1.4 mm apart. The inside area of each plate is 8.8 cm2.

a) What is the capacitance of this set of plates (in pF)

b) If the region between the plates is filled with a material whose dielectric constant is 6.6, what is the new capacitance (in pF)?

2) A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.

a) A voltmeter reads 56.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.9 V. What is the dielectric constant of the material?

b) What will the voltmeter read (in V) if the dielectric is now pulled out so it fills only one-third of the space between the plates?

Answer 1

Given,

d = 1.4 mm = 1.4 x 10^-3 m ; A = 8.8 cm^2 = 8.8 x 10^-4 m^2

a)We know that,

C = e0 A/d

C = 8.85 x 10^-12 x 8.8 x 10^-4/(1.4 x 10^-3) = 5.56 x 10^-12 = 5.56 pF

Hence, C = 5.56 pF

b)for a dielectric with constant k

C = k e0 A/d

C = 6.6 x 8.85 x 10^-12 x 8.8 x 10^-4/(1.4 x 10^-3) = 36.7 x 10^-12 = 36.7 pF

Hence, C = 36.7 pF