Three concentric charged metallic spherical sheets A, B and C have radii a, b, c potentials VA, VB, VC and charge densities +σ, -σ and +σ respectively. The value of potential VA is ( ε₀ = permittivity of free space) (a) (σ)/(ε₀)(a + b + c) (b) (σ)/(ε₀)(-a + b - c) (c) (σ)/(ε₀)(a - b + c) (d) (σ)/(3ε₀)(a + b + c)

Answer: (c). Charges: 4π a²σ, -4π b²σ, 4π c²σ. Inside all shells each adds (q)/(4πε₀ r). VA = (1)/(4πε₀)((4π a²σ)/(a) - (4π b²σ)/(b) + (4π c²σ)/(c)) VA = (σ)/(ε₀)(a - b + c) So the answer is (c).

Three concentric charged metallic spherical sheets A, B and C have… | MHT CET 2025 (20 Apr, Shift 2)

Q37MHT CET 2025 · 20 APR · SHIFT 2Difficult

Electric Potential & Equipotentials

Three concentric charged metallic spherical sheets $A, B$ and $C$ have radii $a, b, c$ potentials $V_A, V_B, V_C$ and charge densities $+\sigma, -\sigma$ and $+\sigma$ respectively. The value of potential $V_A$ is ( $\varepsilon_0 =$ permittivity of free space)

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Answer: (c)

Charges: $4\pi a^2\sigma$ , $-4\pi b^2\sigma$ , $4\pi c^2\sigma$ . Inside all shells each adds $\frac{q}{4\pi\varepsilon_0 r}$ .
$V_A = \frac{1}{4\pi\varepsilon_0}\left(\frac{4\pi a^2\sigma}{a} - \frac{4\pi b^2\sigma}{b} + \frac{4\pi c^2\sigma}{c}\right)$
$V_A = \frac{\sigma}{\varepsilon_0}(a - b + c)$
So the answer is (c).

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Electric Potential & Equipotentials question from MHT CET 2025 (20 Apr, Shift 2)

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