Question

Figure shows two capacitors in series each having area $A$. The rigid central conducting section of length $b$ being movable in vertical direction.


Then, the equivalent capacitance of the given structure is

• A. The equivalent capacitance of the given structure is $\frac{{\epsilon }_{0}A}{(a-b-x)}$
• B. The equivalent capacitance of the given structure is $\frac{{\epsilon }_{0}A}{(a-b)}$
• C. If $x$ becomes $\frac{a-b}{2}$, then equivalent capacitance of the given structure is $\frac{2{\epsilon }_{0}A}{(a-b)}$
• D. If $x$ becomes $\frac{a-b}{2}$, then equivalent capacitance of the given structure is $\frac{{\epsilon }_{0}A}{(a-b)}$

Answer 1

Answer: (B), (D)

The correct options are
B The equivalent capacitance of the given structure is $\frac{{\epsilon }_{0}A}{(a-b)}$
D If $x$ becomes $\frac{a-b}{2}$, then equivalent capacitance of the given structure is $\frac{{\epsilon }_{0}A}{(a-b)}$

The circuit is equivalent to two capacitors in series


$C_1=\frac{{\epsilon }_{0}A}{x}$
$C_2=\frac{{\epsilon }_{0}A}{(a-b-x)}$
$\begin{array}{cc}\frac{1}{C_{eq}}& =\frac{1}{{\epsilon }_{0}A}[a-b-x+x]=\frac{(a-b)}{{\epsilon }_{0}A}\\ \Rightarrow C_{eq}& =\frac{{\epsilon }_{0}A}{(a-b)}\left(\text{independent of}x\right)\end{array}$
Therefore, $C_{eq}$ always remains same irrespective of value of $x$.