A) \[4:1\]
B) \[2:1\]
C) \[1:1\]
D) \[1:2\]
Correct Answer: B
Solution :
: Force between the two charges \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q(q-Q)}{{{r}^{2}}}\] \[\therefore \] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}(Qq-{{Q}^{2}})\] For force to be maximum\[\frac{dF}{dQ}=0\] \[\therefore \]\[\frac{dF}{dQ}=\frac{1}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}.[q-2Q]\] or\[0=q-2Q\] or \[2Q=q\] or \[Q=q/2\] \[\therefore \]\[\frac{q}{q-Q}=\frac{q}{q-\frac{q}{2}}=\frac{q}{q/2}=\frac{2}{1}\]
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