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Question: Answered & Verified by Expert


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
PhysicsElectrostaticsWBJEEWBJEE 2022
Options:
  • 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
Solution:
2111 Upvotes Verified Answer
The correct answer is: $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{a}} \mathrm{q}_{\mathrm{b}}}{\mathrm{c}^2}$
Considering only interaction between charge $q_a$ and $q_b$.

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