Question

Two identical parallel plate air capacitors are connected in series to a battery of emf . If one of the capacitor is completely filled with dielectric material of constant, then potential difference of the other capacitor will become

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

Answer 1

Answer: (A)

Both the capacitance will have same charge $V_{1}=\frac{q}{C}$ and $V_{2}=\frac{q}{K C}$
But $V_{1}+V_{2}=V$
$ \begin{array}{l} \Rightarrow \frac{q}{C}+\frac{q}{K C}=V \\ \Rightarrow \frac{q}{C} \cdot \frac{K+1}{K}=V \end{array} $
Potential difference across capacitor without dielectric is
$ V=\frac{q}{C}=\frac{K}{K+1} V $