StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A thin spherical insulating shell of radius R carries uniformly distributed charge giving a surface potential V0. A small hole of area alpha*4*pi*R² (with alpha << 1) is punched in the shell without disturbing the rest. Which statement is correct? (1) The ratio of the potential at the centre to that at a point R/2 from the centre toward the hole is (1-alpha)/(1-2alpha). (2) The electric field magnitude at the centre is reduced by alpha*V0/(2R). (3) The electric field magnitude at a point on the line through the hole and centre, at distance 2R from the centre, is reduced by alpha*V0/(2R). (4) The potential at the centre is reduced by 2*alpha*V0.

  1. (1)
  2. (2)
  3. (3)
  4. (4)

Correct answer: (2)

Solution

Treat the holed shell as a complete shell plus a small patch of charge -dq located at the hole. For the complete shell the field inside is zero. The missing patch behaves like a small disc of surface charge density sigma. Near the patch its field is sigma/(2*epsilon₀). Using V0 = sigma*R/epsilon₀ => sigma = epsilon₀*V0/R, the field magnitude introduced (the reduction at the centre relative to the now-leaking field) is sigma/(2*epsilon₀) = V0/(2R), and with the fractional area factor alpha it is alpha*V0/(2R). Hence the magnitude of the electric field at the centre is reduced/changed by alpha*V0/(2R). This is the standard JEE result, option (2).

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →