Verified Solution • 2025–26 Session moderate

QJEEAdv19Ph48d4: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph48d4 Solution: A thin spherical insulating shell of radius R carries a…

A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its surface is V0. A hole with a small area α\alpha 4 π\pi R2( α\alpha << 1) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

Multiple Choice Options

  • A The ratio of the potential at the center of the shell of that of the point at 12{1 \over 2} R from center towards the hole will be 1−α1−2α{{1 - \alpha } \over {1 - 2\alpha }} .
  • B The potential at the center of the shell is reduced by 2 α\alpha V0.
  • C The magnitude of electric field at the center of the shell is reduced by αV02R{{\alpha {V_0}} \over {2R}} .
  • D The magnitude of electric field at a point, located on a line passing through the hole and shell's center, on a distance 2R from the center of the spherical shell will be reduced by αV02R{{\alpha {V_0}} \over {2R}} .

Step-by-Step Verified Solution

JEE Advanced 2019 Paper 1 Offline Physics - Electrostatics Question 32 English Explanation

Given, V at surface of the sphere

V0=KQR{V_0} = {{KQ} \over R}

Here, K=14πε0K = {1 \over {4\pi {\varepsilon _0}}} = constant

V at point C,

VC=KQR−KαQR=V0(1−α){V_C} = {{KQ} \over R} - {{K\alpha Q} \over R} = {V_0}(1 - \alpha )

V at point B,

VB=KQR−K(αQ)R/2=V0(1−2α){V_B} = {{KQ} \over R} - {{K(\alpha Q)} \over {R/2}} = {V_0}(1 - 2\alpha )

∴\therefore VCVB=1−α1−2α{{{V_C}} \over {{V_B}}} = {{1 - \alpha } \over {1 - 2\alpha }}

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