A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net electric field E at the centre O is: ![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/3a7e038d-b34f-4e35-ba34-be6842a0978f.png)

![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/5abb9a90-d8fa-4f6b-a43e-c67edeb12274.png) Linear charge density λ = q/π r .Consider an element dl = r dθ at angle θ . dq = λ dl = q/π r (r dθ) = q/π dθ .Electric field due to element dE = 1/4π_0 dq/r^2 .Horizontal components ( dE θ ) cancel out.Vertical components ( dE θ ) add up in the -j direction. E_net = _0^π dE θ (-j) E_net = _0^π 1/4π_0 r^2 q/π θ dθ (-j) E_net = q/4π^2 _0 r^2 [-θ]_0^π (-j) E_net = q/4π^2 _0 r^2 [1 - (-1)]

A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net electric field E at the centre O is: ![image](htt - Physics MCQ AIEEE 2026 | ExamTest.live

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A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net electric field E at the centre O is:

A $\frac{q}{4 π ^{2} ε _{0} r ^{2}} \hat{j}$
B $- \frac{q}{4 π ^{2} ε _{0} r ^{2}} \hat{j}$
C $- \frac{q}{2 π ^{2} ε _{0} r ^{2}} \hat{j}$
D $\frac{q}{2 π ^{2} ε _{0} r ^{2}} \hat{j}$

Solution & Step-by-step Explanation

Linear charge density λ = \frac{q}{π r} .Consider an element d l = r d θ at angle θ . d q = λ d l = \frac{q}{π r} (r d θ) = \frac{q}{π} d θ .Electric field due to element d E = \frac{1}{4 π ε _{0}} \frac{d q}{r ^{2}} .Horizontal components (d E cos θ) cancel out.Vertical components (d E sin θ) add up in the - \hat{j} direction. E_{n e t} = \int_{0}^{π} d E sin θ (- \hat{j}) E_{n e t} = \int_{0}^{π} \frac{1}{4 π ε _{0} r ^{2}} \frac{q}{π} sin θ d θ (- \hat{j}) E_{n e t} = \frac{q}{4 π ^{2} ε _{0} r ^{2}} [- cos θ]_{0}^{π} (- \hat{j}) E_{n e t} = \frac{q}{4 π ^{2} ε _{0} r ^{2}} [1 - (- 1)] (- \hat{j}) = \frac{2 q}{4 π ^{2} ε _{0} r ^{2}} (- \hat{j}) = - \frac{q}{2 π ^{2} ε _{0} r ^{2}} \hat{j}

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A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net electric field E at the centre O is:

\frac{q}{4 π ^{2} ε _{0} r ^{2}} \hat{j}

- \frac{q}{4 π ^{2} ε _{0} r ^{2}} \hat{j}

- \frac{q}{2 π ^{2} ε _{0} r ^{2}} \hat{j}

\frac{q}{2 π ^{2} ε _{0} r ^{2}} \hat{j}

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A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net electric field $E$ at the centre $O$ is:

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