Question

Two parallel plate air filled capacitors, each of capacitance $C$, are joined in series to a battery of emf $V$. The space between the plates of one of the capacitors is then completely filled up with a uniform dielectric having dielectric constant $K$. The quantity of charge which flows through the battery is

• A. $\frac{CV}{2}\left(\frac{K-1}{K+1}\right)$
• B. $\frac{CV}{2}\left(\frac{K+1}{K-1}\right)$
• C. $\frac{2}{CV}\left(\frac{K-1}{K+1}\right)$
• D. $\frac{2}{CV}\left(\frac{K+1}{K-1}\right)$

Answer 1

The correct option is A $\frac{CV}{2}\left(\frac{K-1}{K+1}\right)$

Inital capacitance $=\frac{C}{2}$
Final capacitance $=\frac{KC\times C}{KC+C}=\frac{KC}{K+1}$
Capacitance changes from $\frac{C}{2}$ to $\frac{KC}{K+1}$
Therefore, charge stored in system changes from $\frac{CV}{2}$ to $\frac{KCV}{K+1}$
Quantity of charge flowing through the battery
$=\frac{KCV}{K+1}-\frac{CV}{2}=\frac{CV}{2}[\frac{2K}{K+1}-1]$
$=\frac{CV}{2}\left(\frac{K-1}{K+1}\right)$