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Physics · Electrostatics

JEE Advanced 2025 — Paper 2 — Question 5

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

theradial 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:

  1. Option A:

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

    Correct
  2. Option B:

    Charge flowing from the sphere to the ground because of grounding is 5×10−9C5 \times 10^{-9} \mathrm{C}.

    Correct
  3. Option C:

    After the grounding is removed, the charge on the sphere is −5×10−9C-5 \times 10^{-9} \mathrm{C}.

    Correct
  4. Option D:

    The final electrostatic potential of the sphere is 300 V

Answer: A, B, C

Step-by-step solution

figure

Before grounding Vsphere  ⁣ ⁣  ⁣ ⁣ =(VC)net  ⁣ ⁣  ⁣ ⁣ =(VC)q+(VC)ind  ⁣ ⁣  ⁣ ⁣ {{\text{V}}_{\text{sphere }\!\!~\!\!\text{ }}}={{\left( {{\text{V}}_{\text{C}}} \right)}_{\text{net }\!\!~\!\!\text{ }}}={{\left( {{\text{V}}_{\text{C}}} \right)}_{\text{q}}}+{{\left( {{\text{V}}_{\text{C}}} \right)}_{\text{ind }\!\!~\!\!\text{ }}} Vsphere  ⁣ ⁣  ⁣ ⁣ =kqℓ+0=9×109×10−80.2=900.2=9002=450{{\text{V}}_{\text{sphere }\!\!~\!\!\text{ }}}=\frac{\text{kq}}{\ell }+0=\frac{9\times {{10}^{9}}\times {{10}^{-8}}}{0.2}=\frac{90}{0.2}=\frac{900}{2}=450 volt After grounding

kQℓ+kqsR=V’ sphere  ⁣ ⁣  ⁣ ⁣ ′=0\frac{\text{kQ}}{\ell }+\frac{\text{k}{{\text{q}}_{\text{s}}}}{\text{R}}=\text{{V}'}~_{\text{sphere }\!\!~\!\!\text{ }}^{'}=0 qs=−Rℓq=−12×10−8=−5×10−9{{\text{q}}_{\text{s}}}=-\frac{\text{R}}{\ell }\text{q}=-\frac{1}{2}\times {{10}^{-8}}=-5\times {{10}^{-9}} qs=−5×10−9{{\text{q}}_{\text{s}}}=-5\times {{10}^{-9}} Coulomb Charge flower from sphere to ground =5×10−9=5\times {{10}^{-9}} Coulomb After grounding is removed (Vsphere  ⁣ ⁣  ⁣ ⁣ )final  ⁣ ⁣  ⁣ ⁣ =kqℓ′+kqsR{{\left( {{\text{V}}_{\text{sphere }\!\!~\!\!\text{ }}} \right)}_{\text{final }\!\!~\!\!\text{ }}}=\frac{\text{kq}}{{{\ell }'}}+\frac{\text{k}{{\text{q}}_{\text{s}}}}{\text{R}} =9×109×102×10−830−9×109×5×10−9×10210=\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} =9×100030−450=300volt−450volt=−150volt=\frac{9\times 1000}{30}-450=300\text{volt}-450\text{volt}=-150\text{volt}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Conductors and Redistribution of Charge

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