Physics - Electrostatics Question with Solution | TestHub

Due to charge at $\mathrm{A}$ and $\mathrm{B}$ magnitude of intensity of electric field at point C ${\mathrm{E}}_1={\mathrm{E}}_2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\mathrm{q}}{{\mathrm{r}}^2}$
Net intensity at point $\mathrm{C}$ is $\begin{array}{l}{\mathrm{E}}_{\mathrm{R}}=\sqrt{{\mathrm{E}}_1^2+{\mathrm{E}}_2^2+2{\mathrm{E}}_1{\mathrm{E}}_2\cos 60^{\circ }}\\ =\sqrt{{\mathrm{E}}_1^2+{\mathrm{E}}_1^2+2{\mathrm{E}}_1^2\times \frac{1}{2}}=\sqrt{3}{\mathrm{E}}_1=\frac{\sqrt{3}\mathrm{q}}{4\pi {\epsilon }_0{\mathrm{r}}^2}\end{array}$