The electric potential at the centre of two concentric half rings of radii R ₁ and R ₂, having same linear charge density λ is: (a) (λ)/(2ε₀) (b) (λ)/(ε₀) (c) (2λ)/(ε₀) (d) (λ)/(4ε₀)

Answer: (a). Each half ring: V = (kλπ R)/(R) = (λ)/(4ε₀), independent of radius. V = (λ)/(4ε₀) + (λ)/(4ε₀) = (λ)/(2ε₀) So the answer is (a) (λ)/(2ε₀).

The electric potential at the centre of two concentric half rings of… | JEE Main 2023 (24 Jan, Shift 2)

Q63JEE Main 2023 · 24 JAN · SHIFT 2Medium

Electric Potential & Equipotentials

The electric potential at the centre of two concentric half rings of radii R ${}_1$ and R ${}_2$ , having same linear charge density $\lambda$ is:

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Answer: (a)

Each half ring: $V = \dfrac{k\lambda\pi R}{R} = \dfrac{\lambda}{4\epsilon_0}$ , independent of radius.
$V = \dfrac{\lambda}{4\epsilon_0} + \dfrac{\lambda}{4\epsilon_0} = \dfrac{\lambda}{2\epsilon_0}$
So the answer is (a) $\dfrac{\lambda}{2\epsilon_0}$ .

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