Question

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. ${}^{\prime }{V}^{\prime }$. If one of the capacitor is completely filled with dielectric material of constant ${}^{\prime }{K}^{\prime }$, then potential difference of the other capacitor will become :

• A. $\frac{K}{V(K+1)}$
• B. $\frac{KV}{K+1}$
• C. $\frac{K-1}{KV}$
• D. $\frac{V}{V(K+1)}$

Answer 1

Answer: (B)

$\frac{KV}{K+1}$

Let the capacitance of capacitor $C_2$ be $C$. Thus capacitance of capacitor $C_1$ is $KC$.

Potential across the capacitor $C_2$, $V_2=\frac{C_2}{C_1+C_2}V$

$\therefore$ $V_2=\frac{KC}{C_1+KC}V$

$⟹$ $V_2=\frac{KV}{K+1}$