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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2025 · Paper 2 · Physics · Electrostatics

A positive point charge of 10^ -8 C is kept at a distance of 20 cm...

A positive point charge of 10−8C10^{-8} \mathrm{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 14πϵ0=9×109Nm2/C2\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 (where ϵ0\epsilon_0 is the permittivity of free space), which of the following statements is/are correct: :

Options

  1. ABefore the grounding, the electrostatic potential of the sphere is 450 V .
  2. BCharge flowing from the sphere to the ground because of grounding is 5×10−9C5 \times 10^{-9} \mathrm{C}.
  3. CAfter the grounding is removed, the charge on the sphere is −5×10−9C-5 \times 10^{-9} \mathrm{C}.
  4. DThe final electrostatic potential of the sphere is 300 V .

Correct answer

A. Before the grounding, the electrostatic potential of the sphere is 450 V .

Step-by-step solution

Before grounding Vsphere =(VC)net =(VC)q+(VC)ind Vsphere =kqℓ+0=9×109×10−80.2=900.2=9002=450volt\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 \mathrm{volt}\end{aligned} After grounding\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 =5×10−9=5 \times 10^{-9} Coulomb After grounding is removed (Vsphere )final =kqℓ′+kqsR=9×109×102×10−830−9×109×5×10−9×10210=9×100030−450=300volt−450volt=−150volt\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}
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