A semicircular arc of radius ' R ' is charged uniformly and has charge per unit length as λ. The electric potential V at the centre of curvature is [ ε₀ = permittivity of vacuum] (a) (λ)/(4ε₀) (b) (λ)/(2ε₀) (c) (λ)/(8ε₀) (d) (λ)/(ε₀)

Answer: (a). All charge is at distance R from the centre: V = kq/R with q = λπ R. V = (λπ R)/(4πε₀ R) = (λ)/(4ε₀) So the answer is (a) (λ)/(4ε₀).

A semicircular arc of radius ' R ' is charged uniformly and has… | MHT CET 2026 (19 May, Shift 1)

Q10MHT CET 2026 · 19 MAY · SHIFT 1Medium

Electric Potential & Equipotentials

A semicircular arc of radius ' $R$ ' is charged uniformly and has charge per unit length as $\lambda$ . The electric potential $V$ at the centre of curvature is [ $\varepsilon_0$ = permittivity of vacuum]

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Answer: (a)

All charge is at distance $R$ from the centre: $V = \dfrac{kq}{R}$ with $q = \lambda\pi R$ .
$V = \dfrac{\lambda\pi R}{4\pi\varepsilon_0 R} = \dfrac{\lambda}{4\varepsilon_0}$
So the answer is (a) $\dfrac{\lambda}{4\varepsilon_0}$ .

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Electric Potential & Equipotentials question from MHT CET 2026 (19 May, Shift 1)

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