Question

A parallel-plate capacitor is made by using two square plates with a side length of 1.6 cm, spaced 4.7 mm apart. It is connected to a 12 V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) If the battery is disconnected and then the plates are pulled apart to a separation of 9.4 mm, what are the answers to parts (a) through (d)? (e) The battery is now replaced with an ac power source with 50 V peak amplitude operating at a frequency f = 3000 Hz. Find the rms current in both circuits (i.e., for both values of the separation between the plates).

Answer 1

a) capacitance of a capacitor is given by

$C=\frac{\varepsilon _{o}A}{d}$

plugging in the values we get

C =4.8 x 10^-13f

b) charge

Q =CV

plugging in we get

Q = 5.79 x 10^-9C

c) electric field E is given by

E= V/d

E= 12/4.7 10^-3

E =255.3 N/c

d) $C=\frac{\varepsilon _{o}A}{d}$

C =Co/2 = 2.41 x 10^-13f

charge remains same

electric field remans same.

e) reactance

Xc = 3.3 x10⁸ ohms

Irms = (V/Xc)/sqrt2

Irms =1.07 x 10^-7A

Irms for second case

Irms = 0.54 x10^-7A