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A charge of total amount Q is distributed over two concentric hollow spheres of radii r and R(R>r ) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

Options

  1. A1 4 ₀ (R-r) Q (R^2+r^2 )
  2. B1 4 ₀ (R+r) Q 2 (R^2+r^2 )
  3. C. 1 4 ₀ (R+r) Q (R^2+r^2 . )
  4. D1 4 ₀ (R-r) Q 2 (R^2+r^2 )

Correct answer

C. . 1 4 ₀ (R+r) Q (R^2+r^2 . )

Step-by-step solution

Let q₁ and q₂ be charge on two spheres of radius ' r ' and ' R ' respectively As, q₁+q₂= Q and ₁= ₂ [Surface charge density are equal] q₁ r r^2 = q₂ 4 R^2 So, q₁= Q r^2 R^2+r^2 and q₂= Q R^2 R^2+r^2 Now, potential, V= 1 4 ₀ [ q₁ r + q₂ R ] aligned & = 1 4 ₀ [ Q r R^2+r^2 + Q R R^2+r^2 ] & = Q(R+r) R^2+r^2 1 4 ₀ aligned

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