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

Answer: (b). Element dq = λ r dθ gives dE = (kλ)/(r)dθ; components along the diameter cancel. E = ∫₀π (kλ)/(r)sin θ dθ = (2kλ)/(r) E = (λ)/(2πε₀ r) So the field is (λ)/(2πε₀ r); option (b) appears to be this expression printed upside down.

A uniformly charged semicircular arc of radius ' r ' has linear… | MHT CET 2021 (20 Sep, Shift 2)

Q31MHT CET 2021 · 20 SEP · SHIFT 2Medium

Electric Field & Field Lines

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

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Answer: (b)

Element $dq = \lambda r\,d\theta$ gives $dE = \dfrac{k\lambda}{r}d\theta$ ; components along the diameter cancel.
$E = \displaystyle\int_0^{\pi} \dfrac{k\lambda}{r}\sin\theta\,d\theta = \dfrac{2k\lambda}{r}$
$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
So the field is $\dfrac{\lambda}{2\pi\varepsilon_0 r}$ ; option (b) appears to be this expression printed upside down.

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Electric Field & Field Lines question from MHT CET 2021 (20 Sep, Shift 2)

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