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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

Option 1)
$\frac{q}{2\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

Option 2)
$\frac{q}{4\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

Option 3)
$-\frac{q}{4\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

Option 4)
$-\frac{q}{2\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

Answers (2)

best_answer

As we learnt in

Electric Field Intensity -

$\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}$

- wherein

$\lambda = \tfrac{dq}{dl}$

$dq=\lambda .dl [dl=rd\theta ]$

$dE =\frac{1}{4\pi\:r\varepsilon _{0}}\frac{dq}{r^{2}}$

$=\frac{\lambda rd\theta }{4\pi\:\varepsilon _{o}r^{2}}$

Net electric field at O is

$E=\int ^{\pi}_{o}dE \sin \theta (-\hat{j})=\int ^{\pi}_{o}\frac{\lambda rd\theta \ }{4\pi\:\varepsilon _{o}r^{2}}\sin\theta (-\hat {j})$

$E=-\int ^{\pi}_{o}\frac{q r\sin \theta \ d\theta }{4\pi^{2}\:\varepsilon _{o}r^{3}} \hat{j}$

$=-\frac{q}{4\pi\:^{2}\varepsilon _{o}r^{2}}(-\cos \theta )^{\pi\:}_{o}\hat{j}$

$\therefore E=-\frac{q}{2\pi\:\varepsilon _{o}r^{2}}\hat{j}$

Option 1)

$\frac{q}{2\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

This is incorrect option

Option 2)

$\frac{q}{4\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

This is incorrect option

Option 3)

$-\frac{q}{4\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

This is incorrect option

Option 4)

$-\frac{q}{2\pi ^{2}\varepsilon _{0}r^{2}}\hat{j}$

This is correct option

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