JEE Physics 2025 Question with Solution

Given: A parallel plate capacitor has plate dimensions $l = 3 \, \text{cm}$ and $b = 1 \, \text{cm}$. From the solution working, the plate separation is taken as $d = 3 \, \mu\text{m}$.

Find: Which listed case increases the capacitance by a factor of $10$.

For a parallel plate capacitor,

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

where

$$A = l \times b$$

So the capacitance changes in proportion to

$$\frac{A}{d}$$

The initial area is

$$A = 3 \times 10^{-2} \times 1 \times 10^{-2} = 3 \times 10^{-4} \, \text{m}^2$$

Hence,

$$C_{\text{initial}} = \frac{\varepsilon_0 \cdot 3 \times 10^{-4}}{3 \times 10^{-6}} = 100\varepsilon_0$$

To increase the capacitance by a factor of $10$,

$$C_{\text{new}} = 10C_{\text{initial}} = 1000\varepsilon_0$$

Now compare the listed cases using $C \propto \frac{A}{d}$.

From the extracted solution, the cases labeled C and E satisfy the required factor of increase. Therefore, among the given answer choices, the correct choice is C and E only.

The correct option is A.

Given: Capacitance of a parallel plate capacitor depends on plate area and separation.

Find: The set of modifications that makes capacitance $10$ times the original value.

Use

$$C \propto \frac{lb}{d}$$

Therefore, to get

$$C' = 10C$$

we need

$$\frac{l'b'/d'}{lb/d} = 10$$

the solution explicitly evaluates the candidate cases and concludes that only cases C and E produce the required multiplication factor.

Hence the listed answer choice corresponding to C and E only is A.

Note: the question appears truncated at $d = 3$ and does not include the full set of cases A to E. The answer is therefore grounded in the solution, which identifies the valid cases.