QJEEM25Phfb9f: JEE Main 2025 Physics - Class 12 JEEM25Phfb9f Solution: A parallel plate capacitor of capacitance 1 µF is charged to a…

The energy density ( $u$ ) between the plates of a capacitor is given by the formula:

$$u=\frac{1}{2}{\epsilon }_0{E}^2$$

where $\varepsilon_0$ is the permittivity of free space ( $8.85 \times 10^{-12} \, \text{F/m}$ ) and $E$ is the electric field between the plates. The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:

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

Given:

Capacitance ( $C$ ) = 1 µF = $1 \times 10^{-6} \, \text{F}$

Potential Difference ( $V$ ) = 20 V

Distance ( $d$ ) = 1 µm = $1 \times 10^{-6} \, \text{m}$

First, calculate the electric field $E$ :

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

Now plug this into the formula for energy density:

$$u=\frac{1}{2}\times 8.85\times {10}^{-12}\text{F/m}\times (2\times {10}^7\text{V/m}{)}^2$$

Calculate $u$ :

$$u=\frac{1}{2}\times 8.85\times {10}^{-12}\times 4\times {10}^{14}$$

$$u=2\times 8.85\times {10}^2$$

$$u=1770{\text{J/m}}^3$$

This value is approximately $1.8 \times 10^3 \, \text{J/m}^3$ . Therefore, the correct option is:

Option A: $1.8 \times 10^3 \, \text{J/m}^3$