JEE Physics 2025 Question with Solution

Given: $\epsilon_0$ is the permittivity of free space and $\Phi_E$ is the electric flux.

Find: The dimension of $\epsilon_0 \frac{d\Phi_E}{dt}$.

Electric flux is given by

$$\Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A}$$

So, the dimension of flux is dimension of electric field multiplied by area.

Using dimensional formula of electric field,

$$[E] = \frac{MLT^{-3}}{A}$$

Hence,

$$[\Phi_E] = [E][A] = \frac{MLT^{-3}}{A} \times L^2 = \frac{ML^3T^{-3}}{A}$$

Differentiating with respect to time,

$$\left[\frac{d\Phi_E}{dt}\right] = \frac{ML^3T^{-4}}{A}$$

Now, the dimension of permittivity of free space is

$$[\epsilon_0] = \frac{A^2T^4}{ML^3}$$

Therefore,

$$\left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = \frac{A^2T^4}{ML^3} \times \frac{ML^3T^{-4}}{A} = A$$

Thus, the quantity has the dimension of electric current.

The correct option is D.

Given: $\epsilon_0$ denotes permittivity of free space and $\Phi_E$ denotes electric flux.

Find: The physical quantity having the same dimensions as $\epsilon_0 \frac{d\Phi_E}{dt}$.

From the flux definition,

$$\Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A}$$

The rate of change of electric flux has dimensions of flux per unit time.

From the solution, we use

$$[\epsilon_0] = \frac{A^2T^4}{ML^3}$$

and

$$\left[\frac{d\Phi_E}{dt}\right] = \frac{ML^3T^{-4}}{A}$$

Multiplying,

$$\left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = \frac{A^2T^4}{ML^3} \times \frac{ML^3T^{-4}}{A}$$ $$= A$$

Since $A$ is the dimensional formula of electric current, the required quantity is electric current.

Therefore, the correct option is D.