JEE Physics 2025 Question with Solution

Given: Capacitance of the parallel-plate capacitor is $C = 40 \times 10^{-6} \, \text{F}$, applied voltage is $V = 100 \, \text{V}$, and dielectric constant is $K = 2$.

Find: The extra charge supplied by the battery and the change in electrostatic energy after inserting the dielectric while the power supply remains connected.

When the battery remains connected, the potential difference stays constant. The new capacitance becomes

$$C' = KC = 2 \times 40 \times 10^{-6} = 80 \times 10^{-6} \, \text{F}$$

Initial charge:

$$Q_1 = CV = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3} \, \text{C} = 4 \, \text{mC}$$

Final charge:

$$Q_2 = C'V = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3} \, \text{C} = 8 \, \text{mC}$$

So, the extra charge is

$$\Delta Q = Q_2 - Q_1 = 8 \, \text{mC} - 4 \, \text{mC} = 4 \, \text{mC}$$

Now use the energy formula for constant potential:

$$U = \frac{1}{2}CV^2$$

Initial energy:

$$U_1 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2 \, \text{J}$$

Final energy:

$$U_2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4 \, \text{J}$$

Hence, the change in electrostatic energy is

$$\Delta U = U_2 - U_1 = 0.4 - 0.2 = 0.2 \, \text{J}$$

Therefore, the extra charge is $4\,\text{mC}$ and the increase in electrostatic energy is $0.2\,\text{J}$. The correct option is C.

Given: $C = 40 \times 10^{-6} \, \text{F}$, $V = 100 \, \text{V}$, $K = 2$.

Find: Extra charge and change in electrostatic energy.

The increase in capacitance due to dielectric insertion is

$$\Delta C = C' - C = KC - C = (K-1)C$$

Thus,

$$\Delta Q = \Delta C \cdot V = (KC - C)V$$

Substituting the values,

$$\Delta Q = (2 \times 40 \times 10^{-6} - 40 \times 10^{-6}) \times 100$$ $$= 40 \times 10^{-6} \times 100$$ $$= 4 \times 10^{-3} \, \text{C} = 4 \, \text{mC}$$

For energy change,

$$\Delta U = \frac{1}{2}C'V^2 - \frac{1}{2}CV^2 = \frac{1}{2}(C' - C)V^2 = \frac{1}{2}(K-1)CV^2$$

Now substitute the values,

$$\Delta U = \frac{1}{2}(2-1)(40 \times 10^{-6})(100)^2$$ $$= \frac{1}{2} \times 40 \times 10^{-6} \times 10000$$ $$= 0.2 \, \text{J}$$

Therefore, the required values are $4\,\text{mC}$ and $0.2\,\text{J}$.