Question

A parallel plate capacitor of 2 pF is charged to 12 V and then isolated from the power source. The space between the plates is now filled with a dielectric plate sheet whose dielectric constant is 5.0. The new voltage across the capacitor is

with steps please

Answer 1

Given C = 2pF = 2*10^-12 F

V =12 V

First Find the charge

Q =CV

Q = 2*10^-12*12 = 2.4 *10^-11C

Q= 2.4*10^-11 A

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When a dielectric is inserted the capaitance increases

C_new = kC = 5*C = 10pF

Charge Q remains same ,so new value of voltage across the capacitor is

V =Q/C = 2.4*10^-11/10*10^-12

V =2.4 V

ANSWER: V =2.4 V

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