1 Consider a parallel plate capacitor with plates of area104

1.) Consider a parallel plate capacitor with plates of area1.04 m²whose plates are 1.30 cm apart. Thegap between the plates is filled with air (assume that_air is unity) and thepositive plate has a charge of 11.4 nC on it while the negative plate has anegative charge of equal magnitude on it.

a) What is the electric potential difference (or voltage) acrossthe capacitor?
1 V

The capacitor has its plate-spacing reduced to 1.01 mm, and the plates are connected to a1.55 V battery.

b) What is the charge on the positive plate of thecapacitor?
2 nC

The capacitor is disconnected from the battery and the platesare returned to their initial spacing.

c) What is the voltage across the capacitor?
3 V

The space between the plates is now filled with a fluid with adielectric constant of 4.43.

d) What is the voltage across the capacitor?
4 V

The area of overlap between the capacitor plates is increased to2.31 m².

e) What is the charge on the positive plate of thecapacitor?
5 nC

f) What is the voltage across the capacitor?
6 V

(a)

Capacitance (C) of the capacitor is C = A_o /d

= (1.04 m²)(8.85*10^-12C²/N.m²)/ (1.30*10^-2m)

= 0.708*10^-9F

Charge(Q) across the plates of the capacitor =11.4*10^-9C

Potential difference (V) between the plates of the capacitoris

V = Q / C

= (11.4*10^-9C) / (0.708*10^-9F)

= 16.1V

(b)

In this case the capacitance (C) of the capacitor is

C = A_o / d

= (1.04 m²)(8.85*10^-12C²/N.m²)/ (1.01*10^-3m)

= 9.11nF

Then the charge (Q) across each plate of the capacitoris

Q = CV

= (9.11nF)(1.55V) = 14.12nC

When the dielectric is placed in between the plates of thecapacitor then

V = V_o / k