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Question

A thin semi-circular ring of radius $$R$$ has a positive charge $$q$$ distributed uniformly over it. The net field $$\vec E$$ at the centre $$O$$ is:
1080287_305edbbc96a64f9c8a7fdb87cffd95a3.png


A
q4π2ε0R2^j
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B
q4π2ε0R2^j
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C
q2π2ε0R2^j
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D
q2π2ε0R2^j
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Solution

The correct option is C $$ -\dfrac { q }{ 2{ \pi }^{ 2 }{ \varepsilon }_{ 0 }{ R }^{ 2 } } \hat { j }$$

Given,

$$k=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\,\,and\,\,\lambda =\dfrac{q}{\pi R}\,\,\,\,\,(\,where,\,\lambda =\text{linear}\,\text{charge}\,\text{density})$$

$$\text{Small}\,\text{charge}\,\text{on}\,\text{small}\,\text{length(Rd}\theta )\,\,is\,,\,\,dq=\lambda Rd\theta $$

$$ \vec{E}=\int\limits_{-\pi /2}^{\pi /2}{dE}\cos \theta =2\int\limits_{0}^{\pi /2}{\dfrac{k\left( \lambda Rd\theta  \right)}{{{R}^{2}}}}\cos \theta \,\left( -\hat{j} \right) $$

$$ \Rightarrow \vec{E}=2\times \left( \dfrac{1}{4\pi {{\varepsilon }_{0}}} \right)\,\,\,\left( \dfrac{q}{\pi R} \right)\dfrac{R}{{{R}^{2}}}[\sin \theta ]_{0}^{\pi /2}\,\left( -\hat{j} \right) $$

$$ \Rightarrow \dfrac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}[sin90-\sin 0]\left( -\hat{j} \right) $$

$$ \Rightarrow \vec{E}=\dfrac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}(-\hat{j}) $$

Hence, Electric field at point O is $$\dfrac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}(-\hat{j})$$ 


Physics

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