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)

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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