Question

Two identical parallel plate capacitance C each are connected in series with a battery of emf, E as shown, If one of the capacitors is now filled with a dielectric of dielectric constant k, the amount of charge which will flow through the battery is $?$(neglect internal resistance of the battery)

• A. $\frac{k+1}{2(k-1)}CE$
• B. $\frac{k-1}{2(k+1)}CE$
• C. $\frac{k-2}{k+2}CE$
• D. $\frac{k+2}{k-2}CE$

Answer 1

The correct option is B $\frac{k-1}{2(k+1)}CE$

Inital charge on both the capacitor $Q=\frac{CE}{2}$

final charge$Q_f=\frac{kCE}{K+1}$

charge flow from batter is $Q_f-Q=\frac{k-1}{2(k+1)}CE$