Three +ve charges of equal magnitude ' \(\mathrm{q}\) ' are placed at the vertices of an equilateral…
- By placing a charge \(Q=\left(-\frac{q}{\sqrt{3}}\right)\) at the centroid of the triangle
- By placing a charge \(Q=\left(\frac{q}{\sqrt{3}}\right)\) at the centroid of the triangle
- By placing a charge \(Q=q\) at a distance \(/\) from all the three charges
- By placing a charge \(Q=-q\) above the plane of the triangle at a distance \(l\) from all the three charges
Solution

\(A D=l \cos 30^{\circ}=\frac{l \sqrt{3}}{2}, A O=\frac{2}{3} A D=\frac{1}{\sqrt{3}}\)
\(2\left|\overrightarrow{F_{C A}}\right| \cos 30^{\circ}=\left|\overrightarrow{F_{C O}}\right|\)
\(2 \times \frac{1}{4 \pi \varepsilon_{0}} \times \frac{q^{2}}{l^{2}} \times \frac{\sqrt{3}}{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{(l / \sqrt{3})^{2}}\)
\(\Rightarrow Q=\frac{q}{\sqrt{3}}\)
Asked in: JEE Mains - Electrostatics - Chapter Test