JEE Physics 2023 Question with Solution

Given: Two infinite thin plane sheets each have uniform surface charge density $\sigma$.

Find: The electric field in regions $I$, $II$ and $III$.

For one infinite plane sheet, the magnitude of electric field on either side is

$$E = \frac{\sigma}{2\varepsilon_0}$$

and for a positively charged sheet, the field is directed away from the sheet.

Assuming the right-hand side is along $\hat{n}$:

$$\vec{E}_I = \frac{\sigma}{2\varepsilon_0}(-\hat{n}) + \frac{\sigma}{2\varepsilon_0}(-\hat{n}) = -\frac{\sigma}{\varepsilon_0}\hat{n}$$ $$\vec{E}_{II} = 0$$ $$\vec{E}_{III} = \frac{\sigma}{2\varepsilon_0}(\hat{n}) + \frac{\sigma}{2\varepsilon_0}(\hat{n}) = \frac{\sigma}{\varepsilon_0}\hat{n}$$

Therefore, the field configuration matches option D.

The solution labels the correct option as C, but the extracted working clearly corresponds to option D in the listed options.

Given: Both sheets carry equal positive surface charge density $\sigma$.

Find: Net electric field in each region.

Use the principle of superposition. Each sheet produces field of magnitude

$$\frac{\sigma}{2\varepsilon_0}$$

on both sides.

• In region $I$, both fields point to the left, i.e. along $-\hat{n}$, so they add.
• In region $II$, the field due to left sheet is along $+\hat{n}$ while that due to right sheet is along $-\hat{n}$, so they cancel.
• In region $III$, both fields point along $+\hat{n}$, so they add.

Hence,

$$\vec{E}_I = -\frac{\sigma}{\varepsilon_0}\hat{n}, \qquad \vec{E}_{II} = 0, \qquad \vec{E}_{III} = \frac{\sigma}{\varepsilon_0}\hat{n}$$

Thus, the correct option is D.