Question

A radio capacitor of variable capacitance is made of $n$ parallel plates each of area $A$ and separated from each other by a distance $d$. The alternate plates are connected together. The capacitance of the combination is:

• A. $\frac{nA{ϵ}_0}{d}$
• B. $\frac{(n-1)A{ϵ}_0}{d}$
• C. $\frac{(2n-1)A{ϵ}_0}{d}$
• D. $\frac{(n-2)A{ϵ}_0}{d}$

Answer 1

The correct option is B $\frac{(n-1)A{ϵ}_0}{d}$

Let us assume that the potential of all plates connected from first plate is $V$ and that connected to the second plate is $0$.

So, there are $n-1$ parallel plate capacitors between potentials $V$ and $0$.

$\therefore$ $C_{net}=(n-1)\frac{A{ϵ}_0}{d}$, where $A$ is the area of each plate and $d$ is the separation between each plate.