Question

The figure shows two capacitors in series, the rigid central conducting section of length $b$ being movable vertically. The equivalent capacitance of the given structure is :

• A. $\frac{{\epsilon }_{0}A}{(a-b-x)}$
• B. $\frac{{\epsilon }_{0}A}{(a-b)}$
• C. $\frac{{\epsilon }_{0}A}{(b-x)}$
• D. $\frac{{\epsilon }_{0}A}{(a-x)}$

Answer 1

The correct option is B $\frac{{\epsilon }_{0}A}{(a-b)}$

The circuit is equivalent to two capacitors in series.
$C_1=\frac{{\epsilon }_{0}A}{x}$
$C_2=\frac{{\epsilon }_{0}A}{a-b-x}$
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{{\epsilon }_{0}A}[a-b-x+x]=\frac{(a-b)}{{\epsilon }_{0}A}=\frac{{\epsilon }_{0}A}{(a-b)}$