JEE Physics 2025 Question with Solution

Given: Capacitance $C = 1 \times 10^{-6} \, \text{F}$, potential difference $V = 20 \, \text{V}$, and plate separation $d = 1 \times 10^{-6} \, \text{m}$.

Find: The energy density between the plates of the capacitor.

Use the relation for electric field between parallel plates:

$$E = \frac{V}{d}$$

Substituting the given values:

$$E = \frac{20}{1 \times 10^{-6}} = 2 \times 10^{7} \, \text{V/m}$$

Now use the energy density formula:

$$u = \frac{1}{2} \varepsilon_0 E^2$$

with $\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$. Substitute the values:

$$\nu = \frac{1}{2} \times 8.854 \times 10^{-12} \times (2 \times 10^{7})^2$$ $$\nu = 0.5 \times 8.854 \times 10^{-12} \times 4 \times 10^{14}$$ $$\nu = 17.708 \times 10^{2} = 1.77 \times 10^{3} \, \text{J/m}^3$$

Approximating,

$$\nu \approx 1.8 \times 10^{3} \, \text{J/m}^3$$

Therefore, the correct option is C.

Given: For the capacitor, $V = 20 \, \text{V}$ and $d = 1 \times 10^{-6} \, \text{m}$.

Find: Energy density between the plates.

First identify the electric field using the hint relation:

$$E = \frac{V}{d}$$

So,

$$E = \frac{20}{1 \times 10^{-6}} = 20 \times 10^{6} = 2 \times 10^{7} \, \text{V/m}$$

The energy density stored in the electric field is:

$$U = \frac{1}{2} \epsilon_0 E^2$$

Hence,

$$U = \frac{1}{2} \times 8.854 \times 10^{-12} \times (2 \times 10^{7})^2$$ $$U = 1.77 \times 10^{3} \, \text{J/m}^3$$

Thus the energy density is approximately $1.8 \times 10^{3} \, \text{J/m}^3$, so the correct option is C.