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Question

Physics Question on Electric charges and fields

Charge qq is uniformly distributed over a thin half ring of radius RR. The electric field at the centre of the ring is

A

q2π2ε0R2\frac{ q}{ 2 \pi^2 \varepsilon_0 R^2}

B

q4π2ε0R2\frac{ q}{ 4 \pi^2 \varepsilon_0 R^2}

C

q4πε0R2\frac{ q}{ 4 \pi \varepsilon_0 R^2}

D

q2πε0R2\frac{ q}{ 2 \pi \varepsilon_0 R^2}

Answer

q2π2ε0R2\frac{ q}{ 2 \pi^2 \varepsilon_0 R^2}

Explanation

Solution

From figure di= Rd θ \theta charge on dl = \lambda R d \theta \bigg \\{ \lambda = \frac{ q}{ \pi R } \bigg \\} E electric field at centre due to di is dE = k.λRdθR2\frac{ k . \lambda R d \theta }{ R^2 } We need to consider only the component dE cos θ\theta , as the component dE sin θ\theta will cancel out. Total field at centre = 2∫0π/2dE cos θ 2 \int \limits_0^{ \pi / 2} dE \, cos \, \theta = 2kλR∫0π/2cos dθ\frac{ 2 k \lambda}{ R} \int \limits_0^{ \pi / 2 } cos \, d \theta = 2kλR=q2π2 ε0R2 \frac{ 2 k \lambda}{ R} = \frac{ q}{ 2 \pi^2 \, \varepsilon_0 R^2 }