A) \[q=\frac{Q}{4}\]
B) \[q=\frac{Q}{2}\]
C) \[q=Q\]
D) None of these
Correct Answer: B
Solution :
As total charge \[Eq\] \[{{T}_{1}}=15a\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{15a}{3a}=\frac{5}{1}\] Force between the charges q and q' \[\Rightarrow \] For maximum value of F \[{{T}_{1}}:{{T}_{2}}=5:1\] or \[T=M\,\left( g-\frac{g}{4} \right)=\frac{3Mg}{4}\] \[W=\mathbf{T}\cdot \mathbf{d}\Rightarrow \,W=Td\] \[\Rightarrow \]or \[W=-Td=-\frac{3Mgd}{4}\]
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