JEE Physics 2025 Question with Solution

Given: Mass of bob $m = 100 \times 10^{-6} \, \text{kg}$, charge $q = 10 \times 10^{-6} \, \text{C}$, angle with the sheet $\theta = 45^\circ$, and for an infinite non-conducting sheet the electric field is

Find: The surface charge density $\sigma$.

At equilibrium, the bob is under three forces: weight $mg$ downward, electrostatic force $F_e = qE$ perpendicular to the sheet, and tension along the string. Since the string makes an angle of $45^\circ$, the horizontal and vertical balance gives

and

Dividing the two equations,

Using $F_e = qE = q\frac{\sigma}{2\epsilon_0}$,

So,

Substitute the values from the solution:

Since $\tan 45^\circ = 1$,

This is written in the source solution as $1.77 \, \text{nC/cm}^2$.

Therefore, the correct option is D.

Given: The sheet produces a uniform electric field perpendicular to its surface. The positively charged bob experiences repulsion, so the electric force is horizontal while gravity acts vertically downward.

Find: $\sigma$ using equilibrium of forces.

Resolve tension $T$ into components:

• Horizontal component: $T\sin\theta$
• Vertical component: $T\cos\theta$

Equilibrium requires

and

Now divide the equations to eliminate $T$:

Hence,

Therefore,

Putting $m = 100 \times 10^{-6} \, \text{kg}$, $g = 10 \, \text{m/s}^2$, $\epsilon_0 = 8.85 \times 10^{-12}$, $q = 10 \times 10^{-6} \, \text{C}$, and $\theta = 45^\circ$,

So the answer matches option D as concluded in the solution.