JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2024
- Hard
- MCQ
Two identical conducting spheres \(\mathrm{P}\) and \(\mathrm{S}\) with charge \(Q\) on each, repel each other with a force \(16 \mathrm{~N}\). A third identical uncharged conducting sphere \(\mathrm{R}\) is successively brought in contact with the two spheres. The new force of repulsion between \(\mathrm{P}\) and \(\mathrm{S}\) is what?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(6 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(F_{P S} \propto Q^2\)
\(F_{P S}=16 N\)
Now If \(\mathrm{P} \& \mathrm{R}\) are brought in contact then
Now If \(\mathrm{S} \& \mathrm{R}\) are brought in contact then
New force between \(P \& S\) is :
\(\mathrm{F}_{\mathrm{PS}} \propto \frac{\mathrm{Q}}{2} \times \frac{3 \mathrm{Q}}{4}\)
\(\mathrm{~F}_{\mathrm{PS}} \propto \frac{3 \mathrm{Q}^2}{8}=\frac{3}{8} \times 16=6\)

\(F_{P S}=16 N\)
Now If \(\mathrm{P} \& \mathrm{R}\) are brought in contact then
Now If \(\mathrm{S} \& \mathrm{R}\) are brought in contact then
New force between \(P \& S\) is :
\(\mathrm{F}_{\mathrm{PS}} \propto \frac{\mathrm{Q}}{2} \times \frac{3 \mathrm{Q}}{4}\)
\(\mathrm{~F}_{\mathrm{PS}} \propto \frac{3 \mathrm{Q}^2}{8}=\frac{3}{8} \times 16=6\)

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