JEE Physics 2025 Question with Solution

Given: Two infinitely large parallel conducting plates with surface charge densities $+\sigma$ and $-2\sigma$ are shown, and a point charge $+q$ is placed at the midpoint.

Find: The force on the charge $+q$.

Use superposition of electric fields due to the two infinite plates.

For the left positively charged plate, the field at the midpoint is away from the plate:

$$E_1 = \frac{\sigma}{2\epsilon_0}$$

For the right negatively charged plate of magnitude $2\sigma$, the field at the midpoint is towards the plate:

$$E_2 = \frac{|{-2\sigma}|}{2\epsilon_0} = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$$

Both fields are in the same direction, so

$$E_{\text{net}} = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$$

Hence the force on the positive charge is

$$F = qE_{\text{net}} = q\left(\frac{3\sigma}{2\epsilon_0}\right) = \frac{3q\sigma}{2\epsilon_0}$$

Therefore, the correct option is B and the force is $\frac{3q\sigma}{2\epsilon_0}$, directed towards the plate with charge density $-2\sigma$.

Note: The question text/options mention $-\sigma$, but the solution and diagram clearly use $-2\sigma$. The answer has been derived from the solution, which is the primary source.

Given: A positive sheet on the left and a negative sheet of magnitude $2\sigma$ on the right, with $+q$ at the midpoint.

Find: Net force on $+q$.

Step 1: Electric field due to an infinite sheet has magnitude

$$E = \frac{|\sigma_{\text{sheet}}|}{2\epsilon_0}$$

Step 2: Determine directions.

• Field due to the $+\sigma$ plate points away from that plate, so at the midpoint it is towards the right.
• Field due to the $-2\sigma$ plate points towards that plate, so at the midpoint it is also towards the right.

Step 3: Add magnitudes because both fields are parallel and in the same direction:

$$\begin{aligned} E_{\text{net}} &= \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} \\ &= \frac{3\sigma}{2\epsilon_0} \end{aligned}$$

Step 4: Force on a positive charge has the same direction as the electric field:

$$\vec{F} = q\vec{E}_{\text{net}}$$

So the magnitude is

$$F = \frac{3q\sigma}{2\epsilon_0}$$

Thus, the force acts to the right, towards the negatively charged plate, and the correct option is B.