A parallel-plate capacitor with plate area 2.0 c m 2 and air-gap separation 0.50

Question

A parallel-plate capacitor with plate area2.0cm2and air-gap separation 0.50?mmis connected to a 9.0-Vbattery, and fully charged. The battery is then disconnected.

a) What is the charge on the capacitor?

I got 3.2 x 10^11 C which is correct

b) The plates are now pulled to a separation of 0.75?mm

. What is the charge on the capacitor now?

I got the same thing 3.2 x 10 ^11 C which correct

c) What is the potential difference between the plates now?

Express your answer to two significant figures and include the appropriate units.

d) How much work was required to pull the plates to their new separation?

Explanation / Answer

Q=C1V1=C2V2

C2=(2/3)*C1

V2=3/2 * V1 = 13.5V

d) W= 1/2 *[ C2V2^2 -C1V1^2] = 1/4 * C1V1^2 = (1/4)*(3.2*10^11)*9 =7.2*10^11 J

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