Three identical spheres, each having a change \(q\) and radius \(R\), are kept in such a way that each…
- \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{R}\right)^{2}\)
- \(\frac{\sqrt{3}}{4 \pi \varepsilon_{0}}\left(\frac{q}{R}\right)^{2}\)
- \(\frac{\sqrt{3}}{16 \pi \varepsilon_{0}}\left(\frac{q}{R}\right)^{2}\)
- \(\frac{\sqrt{5}}{16 \pi \varepsilon_{0}}\left(\frac{q}{R}\right)^{2}\)
Solution

Force on \(A\) due to \(B\),
\(F_{A B}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{(2 R)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{4 R^{2}}\) along BA
And force on \(A\) due to \(C\)
\(F_{A C}=\frac{1}{4 \pi \varepsilon_{0}(2 R)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{4 R^{2}} \text { along } \mathrm{CA}\)
Now as angle between \(B A\) and \(C A\) is \(60^{\circ}\) and
$|F_{AB}|=|F_{AC}|=F$ \(\therefore \quad F_{A}=\sqrt{F^{2}+F^{2}+2 F \cdot F \cdot \cos 60}=\sqrt{3} F\)
\(F_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\sqrt{3}}{4}\left(\frac{q}{R}\right)^{2}\)
Asked in: JEE Mains - Electrostatics - Chapter Test