Three concentric spherical shells have radii a, b, and c (a < b< c) and have surface charge densities σ, -σ, and σ respectively. If V_A, V_B and V_C denote the potentials of the three shells, then for c = a + b, we have:

The correct answer is option 2) i.e. VC = VA ≠ VB .

CONCEPT:

• Electric potential due to a conducting spherical shell:
  • Consider a conducting spherical shell of radius R with a charge q on it.
  • To find the electric potential (V) due to the shell at a distance r from the center:

​Case 1: r > R (outside the shell) ⇒           (spherical shell acts similar to point charge)

Case 2: r < R and r = R (inside/surface of the shell) ⇒           (electric field inside the shell is zero and hence electric potential is constant up to a distance R)

CALCULATION:

Given that:

Surface charge density = charge per unit surface area

∴ Charge = surface charge density × surface area

q = σ × A_surafce

Let q_A, q_B, and q_C be the charges on the shell A, B and C respectively.

qA = σ × 4πa²      ----(1)          (∵   Surface area for sphere = 4πr²)

qB = -σ × 4πb2       ----(2) 

qC = σ × 4πc2      ----(3) 

Potential on the surface of shell A:

V_A = 

Substituting (1), (2), and (3) we get

V_A = 

Potential on the surface of shell B:

V_B = 

Substituting (1), (2), and (3) we get

V_B = 

Potential on the surface of shell C:

V_C = 

Substituting (1), (2), and (3) we get

V_C = 

Given that: c = a + b

VA = 

V_B = 

VC = 

Therefore, VC = VA ≠ VB