A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R > r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

Solution

Physics sir · 4 steps

Equal surface density means each shell's charge is proportional to its area, so first find how Q splits.

1. Split the charge. With the same σ on both shells, q₁ = σ · 4π r² and q₂ = σ · 4π R². They must add up to Q, so σ = Q/4π(r² + R²).
2. Potential at centre. The centre is inside both shells. A thin shell gives the same potential everywhere inside it as on its surface, 1/4πε₀q/radius. So the two contributions just add.

   V = 1/4πε₀((σ 4π r²)/r + (σ 4π R²)/R) = 1/4πε₀ 4πσ(r + R)

3. Put in sigma. Using 4πσ = Q/(r² + R²) gives V = 1/4πε₀(R + r)/(R² + r²)Q.
4. So. The potential at the common centre matches option (a).

The idea to keep. Inside a charged shell the potential is constant and equals its surface value kq/R. For several concentric shells, just add each shell's kq/R at any point inside all of them.

Answer from the NTA final answer key. Solution by PrepWiser.