WBJEE · Physics · Electrostatics
A neutral conducting solid sphere of radius \(\mathrm{R}\) has two spherical cavities of radius a and \(b\) as shown in the figure. Centre to centre distance between two cavities is \(c . q_a\) and \(q_b\) charges are placed at the centres of cavities respectively. The force between \(q_a\) and \(q_b\) is
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{a}} \mathrm{q}_{\mathrm{b}}}{\mathrm{c}^2}\)
- B \(\frac{1}{4 \pi \varepsilon_0} q_a q_b\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
- C zero
- D insufficient data
Answer & Solution
Correct Answer
(A) \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{a}} \mathrm{q}_{\mathrm{b}}}{\mathrm{c}^2}\)
Step-by-step Solution
Detailed explanation
Considering only interaction between charge \(q_a\) and \(q_b\).
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