MHT CET Medical202623 April 2026Morning ShiftPhysicsElectrostaticsActual
A charge Q is distributed over the two concentric hollow spheres of radii r and R ( R > r ) such the surface densities are equal. The potential at the common centre is K = Q 4 ₀ , ₀ = permittivity of free space
Options
- AKR r
- BK(r+R) (R^2+r^2)
- CK(R^2+r^2) (R+r)
- DK(r+R) (R^3+r^3)
Correct answer
B. K(r+R) (R^2+r^2)
Step-by-step solution
Let q₁ and q₂ be the charges on the inner and outer spheres respectively. Given q₁ + q₂ = Q Since the surface charge densities are equal, ₁ = ₂ = q₁ 4 r^2 = q₂ 4 R^2 = q₁ = 4 r^2 and q₂ = 4 R^2 Total charge Q = 4 (r^2 + R^2) = Q 4 (r^2 + R^2) The potential at the common centre is the sum of the potentials due to both spheres: V = 1 4 ₀ q₁ r + 1 4 ₀ q₂ R V = 1 4 ₀ ( 4 r^2 r + 4 R^2 R ) V = ₀ (r + R) Substituting the value of : V = Q 4 ₀ (r^2 + R^2) (r + R) Given K = Q 4 ₀ , we get: V = K(r+R) R^2+r^2 Answer: K(r+R)