Question

A capacitor of capacitance $10\mu F$ is connected to battery of emf $20V$. Without disconnecting the source a dielectric $(K=4)$ is introduced to fill the space between two plates of the capacitor. Calculate the-
(a) charge before the dielectric was introduced.
(b) charge After the dielectric was introduced.

Answer 1

Capacitance before introduce dielectric $C_b$

$C_b=10\times {10}^{-6}F$

$Q=C_{b}V={10}^{-5}\times 20=2\times {10}^{-4}C$

Capacitor after introduce dielectric $C_a$

$C_a=k{C}_b=4\times 10\times {10}^{-6}F=4\times {10}^{-5}F$

$Q=C_{a}V=4\times {10}^{-5}\times 20=8\times {10}^{-4}F$

(a) charge before the dielectric was introduced. $Q=2\times {10}^{-4}C$
(b) charge After the dielectric was introduced, $Q=4\times {10}^{-4}C$