Question

A parallel-plate capacitor with circular plates and a capacitance of 10.3 μF is connected to a battery which provides a voltage of 11.2 V .

What is the charge on each plate?

How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery?

How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation?

Answer 1

Part A.

Charge on capacitor plate is given by:

Q = C*V

C = 10.3*10^-6 F & V = 11.2 V, So

Q = 10.3*10^-6*11.2

Q = 115.36*10^-6 C = 115.36 C = Charge on each plate

Part B.

Since battery is still connected, So Voltage across capacitor will remain constant, Now

Q = C*V

When plate separation is doubled, then Capacitance will be halved, (Since C = e0*A/d), So

New Capacitance = C1 = C/2 = (1/2)*10.3*10^-6 F

Now new charge will be, since V is constant, So

Q1 = C1*V = (1/2)*C*V = (1/2)*Q

Q1 = (1/2)*115.36*10^-6 C

Q1 = 57.68*10^-6 C = 57.68 C

Part C.

When radius of each plate is doubled, then C1 = e0*A1/d = e0*(pi*r1^2)/d

Since r1 = 2*r, So

C1 = e0*(pi*(2*r)^2)/d = 4*(e0*pi*r^2/d) = 4*C

Now Since voltage is constant, So

Q1 = C1*V

Q1 = 4*C*V = 4*115.36*10^-6

Q1 = 461.44*10^-6 C = 461.44 C = Charge on each plate