NEET · Physics · STD 12 - 1. Electric charges and fields
Two identical charged conducting spheres \(A\) and \(B\) have their centres separated by a certain distance. Charge on each sphere is \(q\) and the force of repulsion between them is \(F\). A third identical uncharged conducting sphere is brought in contact with sphere \(A\) first and then with \(B\) and finally removed from both. New force of repulsion between spheres \(A\) and \(B\) (Radii of \(A\) and \(B\) are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
- A \(\frac{3 F}{5}\)
- B \(\frac{2 F}{3}\)
- C \(\frac{F}{2}\)
- D \(\frac{3 F}{8}\)
Answer & Solution
Correct Answer
(D) \(\frac{3 F}{8}\)
Step-by-step Solution
Detailed explanation
\( F=\frac{kQ^2}{r^2} \)
on touching; \(F^{\prime}=\frac{k(Q / 2)(3 Q / 4)}{r^2}\)
\( F^{\prime}=\frac{3 F}{8} \)
After touching \(\rightarrow\)
\(( A ) \& ( C ) \rightarrow\)
Total charge \(\rightarrow q +0= q\)
\(Q_A^{\prime}=\frac{q}{2}, \quad Q_C^{\prime}=\frac{q}{2}\)
(B) & (C)
Total charge \(=q+\frac{q}{2}=\frac{3}{2} q\)
\(Q_B^{\prime}=\frac{3}{4} q \quad Q_C^{\prime}=\frac{3}{4} q\)
on touching; \(F^{\prime}=\frac{k(Q / 2)(3 Q / 4)}{r^2}\)
\( F^{\prime}=\frac{3 F}{8} \)
After touching \(\rightarrow\)
\(( A ) \& ( C ) \rightarrow\)
Total charge \(\rightarrow q +0= q\)
\(Q_A^{\prime}=\frac{q}{2}, \quad Q_C^{\prime}=\frac{q}{2}\)
(B) & (C)
Total charge \(=q+\frac{q}{2}=\frac{3}{2} q\)
\(Q_B^{\prime}=\frac{3}{4} q \quad Q_C^{\prime}=\frac{3}{4} q\)
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