JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2026
- Easy
- MCQ
Consider two identical metallic spheres of radius \(R\) each having charge \(Q\) and mass \(m\). Their centers have an initial separation of \(4 R\). Both the spheres are given an initial speed of \(u\) towards each other. The minimum value of \(u\), so that they can just touch each other is:
(Take \(k=\frac{1}{4 \pi \epsilon_0}\) and assume \(k Q^2>G m^2\) where G is the Gravitational constant)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{k Q^2}{4 m R}\left(1-\frac{G m^2}{k Q^2}\right)}\)
Step-by-step Solution
Detailed explanation
Using energy conservation
(2) \(\left(\frac{1}{2} mu ^2\right)-\frac{ Gm ^2}{4 r }+\frac{ KQ ^2}{4 r }=-\frac{ Gm ^2}{2 r }+\frac{ KQ ^2}{2 r }\)
\(u =\sqrt{\frac{1}{4 mr }\left( KQ ^2- Gm ^2\right)}\)
(2) \(\left(\frac{1}{2} mu ^2\right)-\frac{ Gm ^2}{4 r }+\frac{ KQ ^2}{4 r }=-\frac{ Gm ^2}{2 r }+\frac{ KQ ^2}{2 r }\)
\(u =\sqrt{\frac{1}{4 mr }\left( KQ ^2- Gm ^2\right)}\)
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