1.) Consider a parallel plate capacitor with plates of area1.04 m 2 whose plates

Question

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 thatair 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

Explanation / Answer

(a) Capacitance (C) of the capacitor is C = Ao /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 = Ao / 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 = Vo / k                          = (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 = Vo / k

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