Question

Two identical parallel plate capacitors of capacitance C each are connected in series with a battery of Emf, E as shown. If one of the capacitors is now filled with dielectric constant k, the amount of charge which will flow through the battery is

• 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 $\frac{(k-1)}{2(k+1)}CE$

$Q_1=\left(\frac{C}{2}\right)E$
$Q_2=\frac{k}{k+1}CE$
$\Delta Q=Q_2-Q_1$
$=\frac{kCE}{k+1}-\frac{C}{2}E$
$=CE\left(\frac{k-1}{2(k+1)}\right)$