Sankalp NEET Full Test-5 Question-15 Solution

Question: 15 : The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to:

(1) 2 C

(2) \mathrm{C} / 2

(3) 4 C

(4) C / 4

Answer: Option (4)

Explanation:

The capacitance of a parallel plate capacitor is given by

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

where A is the area of each plate and d is the separation between the plates.

Initially, the capacitance is

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

According to the question, the new area of each plate becomes

A' = \frac{A}{2}

and the new separation between the plates becomes

d' = 2d

The final capacitance is

C' = \frac{\varepsilon_0 A'}{d'}

Substituting the new values,

C' = \frac{\varepsilon_0 \left(\frac{A}{2}\right)}{2d}

Simplifying,

C' = \frac{\varepsilon_0 A}{4d}

Since \frac{\varepsilon_0 A}{d} = C, we get

C' = \frac{C}{4}

Therefore, the final capacitance is equal to C/4.

Hence, the correct option is (4).