JEE Advanced · Physics · 17. Electrostatics
Consider a system of three charges \(\frac{q}{3}, \frac{q}{3}\) and \(-\frac{2 q}{3}\) placed at points \(A, B\) and \(C\), respectively, as shown in the figure. Take \(O\) to be the centre of the circle of radius \(R\) and angle \(C A B=60^{\circ}\).

- A The electric field at point \(O\) is \(\frac{q}{8 \pi \varepsilon_0 R^2}\) directed along the negative \(x\)-axis
- B The potential energy of the system is zero
- C The magnitude of the force between the charges at \(C\) and \(B\) is \(\frac{q^2}{54 \pi \varepsilon_0 R^2}\)
- D The potential at point \(O\) is \(\frac{q}{12 \pi \varepsilon_0 R}\)
Answer & Solution
Correct Answer
(C) The magnitude of the force between the charges at \(C\) and \(B\) is \(\frac{q^2}{54 \pi \varepsilon_0 R^2}\)
Step-by-step Solution
Detailed explanation
Distance \(B C=A B \sin 60^{\circ}=(2 R) \frac{\sqrt{3}}{2}=\sqrt{3} R\)
\(\therefore \quad\left|F_{B C}\right|=\frac{1}{4 \pi \varepsilon_0} \frac{(q / 3)(2 q / 3)}{(\sqrt{3} R)^2}=\frac{q^2}{54 \pi \varepsilon_0 R^2}\)
\(\therefore\) correct option is (c).
\(\therefore \quad\left|F_{B C}\right|=\frac{1}{4 \pi \varepsilon_0} \frac{(q / 3)(2 q / 3)}{(\sqrt{3} R)^2}=\frac{q^2}{54 \pi \varepsilon_0 R^2}\)
\(\therefore\) correct option is (c).
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