Question

The capacity of a parallel plate condenser is 12 $\mu$F. Its capacity, when the separation between plates is doubled and area is halved, will be

• A. 3$\mu$F
• B. 12 $\mu$F
• C. 6$\mu$F
• D. 1.5 $\mu$F

Answer 1

Answer: (A)

3 μF

Answer 2

For a parallel‐plate capacitor, the capacitance is given by

$C=\varepsilon_0\frac{A}{d}$.

Initially, the capacitance is $C=12\,\mu F$. Now the area is halved $\left(A'=\frac{A}{2}\right)$ and the plate separation is doubled $\left(d' =2d\right)$. Then the new capacitance is

$C'=\varepsilon_0\frac{A'}{d'}=\varepsilon_0\frac{A/2}{2d}=\frac{\varepsilon_0 A}{4d}=\frac{C}{4}$.

Thus,

$C'=\frac{12\,\mu F}{4}=3\,\mu F$.

So the answer is 3 μF (option a).