JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Two identical conducting spheres \(A\) and \(B,\) carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is \(F\) . A third identical conducting sphere, \(C,\) is uncharged. Sphere \(C\) is first touched to \(A,\) then to \(B,\) and then removed. As a result, the force between \(A\) and \(B\) would be equal to
- A \(\frac{{3F}}{4}\)
- B \(\frac{{F}}{2}\)
- C \(F\)
- D \(\frac{{3F}}{8}\)
Answer & Solution
Correct Answer
(D) \(\frac{{3F}}{8}\)
Step-by-step Solution
Detailed explanation
Spheres \(A\) and \(B\) carry equal charge say \('q\) \(\therefore \) Force between them, \(\mathrm{F}=\frac{\mathrm{k} \mathrm{qq}}{\mathrm{r}^{2}}\) When \(A\) and \(C\) are touched, charge on both \(\mathrm{q}_{\mathrm{A}}=\mathrm{q}_{\mathrm{C}}=\frac{\mathrm{q}}{2}\) Then…
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