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Electrostatics question

2020 · 6 Sep · Shift 2 · Q55
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Electrostatics question

2020 · 6 Sep · Shift 2 · Q55

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell?
  1. A
    14πε0qQR2>F>0{1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{R^2}}} \gt F \gt 04πε0​1​R2qQ​>F>0 for r < R
  2. B
    F=14πε0qQr2F = {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{r^2}}}F=4πε0​1​r2qQ​ for r > R
  3. C
    F=14πε0qQr2F = {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{r^2}}}F=4πε0​1​r2qQ​ for all r
  4. D
    F=14πε0qQR2F = {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{R^2}}}F=4πε0​1​R2qQ​ for r < R
View written solutionFree

Correct answer: B

  1. Use the result for a uniformly charged spherical shell

A spherical shell with total charge QQQ uniformly distributed over its surface has the following electric field:

  • Inside the shell (r<R)(r<R)(r<R): E=0E=0E=0
  • Outside the shell (r>R)(r>R)(r>R): E=14πε0Qr2E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}E=4πε0​1​r2Q​

This follows from Gauss's law.

  1. Force on charge qqq

The force on a test charge qqq is F=qEF=qEF=qE So,

  • For r<Rr<Rr<R: F=0F=0F=0
  • For r>Rr>Rr>R: F=14πε0qQr2F=\frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2}F=4πε0​1​r2qQ​
  1. Check each option

  • Option A: 14πε0qQR2>F>0for r<R\frac{1}{4\pi\varepsilon_0}\frac{qQ}{R^2}>F>0 \quad \text{for } r<R4πε0​1​R2qQ​>F>0for r<R This is false because for r<Rr<Rr<R, actually F=0F=0F=0 not a positive value.

  • Option B: F=14πε0qQr2for r>RF=\frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2} \quad \text{for } r>RF=4πε0​1​r2qQ​for r>R This is true.

  • Option C: F=14πε0qQr2for all rF=\frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2} \quad \text{for all } rF=4πε0​1​r2qQ​for all r False, because for r<Rr<Rr<R, force is zero.

  • Option D: F=14πε0qQR2for r<RF=\frac{1}{4\pi\varepsilon_0}\frac{qQ}{R^2} \quad \text{for } r<RF=4πε0​1​R2qQ​for r<R False, because for r<Rr<Rr<R, force is zero.

  1. Final answer

The correct statement is: B\boxed{B}B​

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