Eleven equal point charges,each of magnitude | vedclass

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Question

(A) If there were $12$ equal charges $+Q$ at all hour positions,the electric field at the centre would be zero due to symmetry,as each charge would be

Vedclass · Accepted Answer

$\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the centre towards the mark $10$

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(A) If there were $12$ equal charges $+Q$ at all hour positions,the electric field at the centre would be zero due to symmetry,as each charge would be cancelled by an equal and opposite charge at the diametrically opposite position.Let $\vec{E}_{10}$ be the electric field at the centre due to a charge $+Q$ at the $10$ o'clock position. The net electric field $\vec{E}_{net}$ with the charge at $10$ missing is given by:$\vec{E}_{total} = \vec{E}_{net} + \vec{E}_{10} = 0$Therefore,$\vec{E}_{net} = -\vec{E}_{10}$.The magnitude of the electric field due to a single charge $+Q$ at distance $r$ is $E = \frac{Q}{4 \pi \varepsilon_{0} r^{2}}$.The direction of $\vec{E}_{10}$ is from the $10$ o'clock position towards the centre.Thus,$-\vec{E}_{10}$ is a vector of magnitude $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ directed from the centre towards the $10$ o'clock position.

Eleven equal point charges,each of magnitude $+Q$,are placed at all the hour positions of a circular clock of radius $r$,except at the $10$ o'clock position. What is the electric field strength at the centre of the clock?

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