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  • A point charge is placed at the Centre of a uniformly charge ring of total charge <img alt="Q" src="https://entrancecorner.on

A point charge q is placed at the Centre of a uniformly charge ring of total charge Q.Two small pieces, each with charge q^{\prime} are removed from the ring as shown in the figure. The net force on the charge q due to the remaining ring.

Option: 1

\frac{1}{4 \pi \epsilon_{0}} \frac{q q^{\prime}}{R^{2}}


Option: 2

\sqrt{2} \times \frac{1}{4 \pi \epsilon_{0}} \frac{q q^{\prime}}{R^{2}}


Option: 3

2 \sqrt{2} \times \frac{1}{4 \pi \epsilon_{0}} \frac{q q_{1}}{R^{2}}


Option: 4

\sqrt{2} \times \frac{3}{4 \pi \epsilon_{0}} \frac{q q^{\prime}}{R^{2}}


Answers (1)

best_answer

If \vec{F}_{1} is force due to remaining ring and \vec{F}_{2}  by the removed charges then -

\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3} =0
\vec{F}_{1} =-\left(\vec{F}_{2}+\vec{F}_{3}\right)
\vec{F}_{1} =\sqrt{F_{2}^{2}+F_{3}^{2}}=\sqrt{F^{2}+F^{2}}

=\sqrt{2 F^{2}}
F_{1} =\sqrt{2} F=\sqrt{2} \times \frac{1}{4 \pi _{0}\epsilon_{0}} \frac{q q^{\prime}}{R^{2}} ans.

Posted by

Gautam harsolia

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