JEE Main 2025 — Electrostatics Question with Solution

Step 1: Balancing Forces on the Bob

At equilibrium, the electric force ($qE$) pulling the bob toward the sheet is balanced by the weight of the bob ($mg$) pulling it down.

Step 2: Formula for Electric Field Near Sheet

The electric field ($E$) created by a large sheet of charge is given by $E=\frac{\sigma }{2{\epsilon }_0}$.

Step 3: Replace $E$ in the Force Equation

The equation from earlier becomes: $$q\left[\frac{\sigma }{2{\epsilon }_0}\right]=mg$$

Step 4: Solve for $\sigma$

We solve for $\sigma$ by rearranging: $$\sigma =\frac{2{\epsilon }_{0}mg}{q}$$

Step 5: Substitute the Values

Put in the given values: ${\epsilon }_0=8.85\times 10^{-12}$ F/m, $m=100$ mg $=100\times 10^{-6}$ kg, $g=10$ m/s${}^2$, $q=10\times 10^{-6}$ C.

So, $$\sigma =\frac{2\times 8.85\times 10^{-12}\times 100\times 10^{-6}\times 10}{10\times 10^{-6}}$$

Step 6: Calculate

We do the multiplication and division: $$\sigma =17.7\times 10^{-10}\mathrm{C}/{\mathrm{m}}^2$$

Step 7: Convert to nC/m${}^2$

$1\mathrm{nC}=10^{-9}\mathrm{C}$, so $$\sigma =1.77\mathrm{nC}/{\mathrm{m}}^2$$