JEE Advance - Physics (2025 - Paper 2 Online - No. 5)

A positive point charge of $10^{-8}$ C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ Nm$^2$/C$^2$ (where $\epsilon_0$ is the permittivity of free space), which of the following statements is/are correct:
Before the grounding, the electrostatic potential of the sphere is 450 V.
Charge flowing from the sphere to the ground because of grounding is $5 \times 10^{-9}$ C.
After the grounding is removed, the charge on the sphere is $-5 \times 10^{-9}$ C.
The final electrostatic potential of the sphere is 300 V.

Explanation

JEE Advanced 2025 Paper 2 Online Physics - Electrostatics Question 2 English Explanation 1

Before grounding

$$ \begin{aligned} & \mathrm{V}_{\text {sphere }}=\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {net }}=\left(\mathrm{V}_{\mathrm{C}}\right)_{\mathrm{q}}+\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {ind }} \\ & \mathrm{V}_{\text {sphere }}=\frac{\mathrm{kq}}{\ell}+0=\frac{9 \times 10^9 \times 10^{-8}}{0.2}=\frac{90}{0.2}=\frac{900}{2}=450 \text { volt } \end{aligned} $$

After grounding

JEE Advanced 2025 Paper 2 Online Physics - Electrostatics Question 2 English Explanation 2

$$ \begin{aligned} & \frac{\mathrm{kQ}}{\ell}+\frac{\mathrm{kq}_{\mathrm{s}}}{\mathrm{R}}=\mathrm{V}_{\text {sphere }}^{\prime}=0 \\ & \mathrm{q}_{\mathrm{s}}=-\frac{\mathrm{R}}{\ell} \mathrm{q}=-\frac{1}{2} \times 10^{-8}=-5 \times 10^{-9} \\ & \mathrm{q}_{\mathrm{s}}=-5 \times 10^{-9} \text { Coulomb } \end{aligned} $$

Charge flower from sphere to ground $=5 \times 10^{-9}$ Coulomb

After grounding is removed

$$ \begin{aligned} & \left(\mathrm{V}_{\text {sphere }}\right)_{\text {final }}=\frac{\mathrm{kq}}{\ell^{\prime}}+\frac{\mathrm{kq}_{\mathrm{s}}}{\mathrm{R}} \\ & =\frac{9 \times 10^9 \times 10^2 \times 10^{-8}}{30}-\frac{9 \times 10^9 \times 5 \times 10^{-9} \times 10^2}{10} \\ & =\frac{9 \times 1000}{30}-450=300 \mathrm{volt}-450 \mathrm{volt}=-150 \mathrm{volt} \end{aligned} $$

Comments (0)

Advertisement