A capacitor of capacitance $10 μ F$ is connected to battery of emf $20 V$ . 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.

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Solution

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, $Q = 4 \times 10^{- 4} C$

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