A uniformly charged semicircular arc of radius ' r ' has linear charge density ' λ '. The electric field at its centre is ( ε₀ = permittivity of free space) (a) (λ)/(4ε₀) (b) (2ε₀)/(λ) (c) (λ)/(4ε₀ r) (d) (2πε₀)/(λ)

Answer: (b). Element dq = λ r dθ gives dE = (kλ r dθ)/(r²); components along the diameter cancel. E = ∫₀π (kλ)/(r)sin θ dθ = (2kλ)/(r) E = (2λ)/(4πε₀ r) = (λ)/(2πε₀ r) So the field is (λ)/(2πε₀ r); none of the printed options matches this exactly.

A uniformly charged semicircular arc of radius ' r ' has linear… | MHT CET 2023 (14 May, Shift 1)

Q22MHT CET 2023 · 14 MAY · SHIFT 1Medium

Electric Field & Field Lines

A uniformly charged semicircular arc of radius ' $r$ ' has linear charge density ' $\lambda$ '. The electric field at its centre is ( $\varepsilon_0 =$ permittivity of free space)

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Answer: (b)

Element $dq = \lambda r\,d\theta$ gives $dE = \dfrac{k\lambda r\,d\theta}{r^2}$ ; components along the diameter cancel.
$E = \displaystyle\int_0^{\pi} \dfrac{k\lambda}{r}\sin\theta\,d\theta = \dfrac{2k\lambda}{r}$
$E = \dfrac{2\lambda}{4\pi\varepsilon_0 r} = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
So the field is $\dfrac{\lambda}{2\pi\varepsilon_0 r}$ ; none of the printed options matches this exactly.

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Electric Field & Field Lines question from MHT CET 2023 (14 May, Shift 1)

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