Six point charges are kept 60^{circ} apart fr | vedclass

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(B) Let the magnitude of the electric field due to a single charge $Q$ at the center be $E_{0} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{R^{2}}$.By analyz

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$ -\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{i}-\hat{j}) $

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(B) Let the magnitude of the electric field due to a single charge $Q$ at the center be $E_{0} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{R^{2}}$.By analyzing the symmetry, the charges at $90^{\circ}$ and $270^{\circ}$ (both $+Q$) cancel each other out.The remaining charges are at $30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}$.Specifically, the charges at $30^{\circ}$ and $210^{\circ}$ are $+Q$ and $+Q$, and at $150^{\circ}$ and $330^{\circ}$ are $-Q$ and $+Q$.Using the principle of superposition, the net electric field is the vector sum of individual fields.After resolving the components, the net electric field at the center is calculated as $\vec{E}_{net} = -\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{i}-\hat{j})$.

Six point charges are kept $60^{\circ}$ apart from each other on the circumference of a circle of radius $R$ as shown in the figure. The net electric field at the centre of the circle is . . . . . . . ($ \epsilon_{0} $ is the permittivity of free space)

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