Eleven equal point charges, each of magnitude | vedclass

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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

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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