Question

Figure shows two capacitors in series, the rigid and neutral central conducting section of length $b$ being movable vertically.


If the potentials of upper and lower plates are $V_1$ and $V_2$ respectively, then the potential of the middle rigid section is

• A. $V_1-\frac{(V_1-V_2)x}{(a-b)}$
• B. $V_1-\frac{(V_2-V_1)x}{(a-b)}$
• C. $V_1-\frac{(V_1-V_2)x}{(a+b)}$
• D. $V_1-\frac{(V_2-V_1)x}{(a+b)}$

Answer 1

The correct option is A $V_1-\frac{(V_1-V_2)x}{(a-b)}$

The capacitance of the induvidual capacitor can be written as
$C_1=\frac{{\epsilon }_{0}A}{x}$
$C_2=\frac{{\epsilon }_{0}A}{a-b-x}$
Since the capacitors are in series, equivalent capacitance will be
$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

Now potential will be distributed between these capacitors.
We can reduce this structure to the following circuit.


Since the charge will be same on the individual capacitor and also on the equivalent capacitor.
Therefore equating the charge on $C_1$ and $C_{eq}$
$(V_1-V)C_1=(V_1-V_2)\frac{C_1{C}_2}{C_1+C_2}$

$V_1-V=(V_1-V_2)\frac{\frac{{\epsilon }_{0}A}{a-b-x}}{\frac{{\epsilon }_{0}A}{x}+\frac{{\epsilon }_{0}A}{a-b-x}}=(V_1-V_2)\frac{x}{(a-b)}$


$V=V_1-(V_1-V_2)\frac{x}{(a-b)}$