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Equal charges $q$ each are placed at the vertices of an equilateral triangle of side $r$. The magnitude of electric field intensity at any vertex is
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Verified Answer
The correct answer is:
$\frac{\sqrt{3} q}{4 \pi \varepsilon_{0} r^{2}}$
Due to charge at $\mathrm{A}$ and $\mathrm{B}$ magnitude of intensity of electric field at point C
$\mathrm{E}_{1}=\mathrm{E}_{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}}$
Net intensity at point $\mathrm{C}$ is
$\begin{array}{l}
\mathrm{E}_{\mathrm{R}}=\sqrt{\mathrm{E}_{1}^{2}+\mathrm{E}_{2}^{2}+2 \mathrm{E}_{1} \mathrm{E}_{2} \cos 60^{\circ}} \\
=\sqrt{\mathrm{E}_{1}^{2}+\mathrm{E}_{1}^{2}+2 \mathrm{E}_{1}^{2} \times \frac{1}{2}}=\sqrt{3} \mathrm{E}_{1}=\frac{\sqrt{3} \mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}
\end{array}$
$\mathrm{E}_{1}=\mathrm{E}_{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}}$
Net intensity at point $\mathrm{C}$ is
$\begin{array}{l}
\mathrm{E}_{\mathrm{R}}=\sqrt{\mathrm{E}_{1}^{2}+\mathrm{E}_{2}^{2}+2 \mathrm{E}_{1} \mathrm{E}_{2} \cos 60^{\circ}} \\
=\sqrt{\mathrm{E}_{1}^{2}+\mathrm{E}_{1}^{2}+2 \mathrm{E}_{1}^{2} \times \frac{1}{2}}=\sqrt{3} \mathrm{E}_{1}=\frac{\sqrt{3} \mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}
\end{array}$
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