A) \[{{Q}_{2}}=\frac{Q}{R},\ {{Q}_{1}}=Q-\frac{Q}{R}\]
B) \[{{Q}_{2}}=\frac{Q}{4},\ {{Q}_{1}}=Q-\frac{2Q}{3}\]
C) \[{{Q}_{2}}=\frac{Q}{4},\ {{Q}_{1}}=\frac{3Q}{4}\]
D) \[{{Q}_{1}}=\frac{Q}{2},\ {{Q}_{2}}=\frac{Q}{2}\]
Correct Answer: C
Solution :
\[{{Q}_{1}}+{{Q}_{2}}=Q\] ..... (i) and \[F=k\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}\] .....(ii) From (i) and (ii) \[F=\frac{k{{Q}_{1}}(Q-{{Q}_{1}})}{{{r}^{2}}}\] For F to be maximum \[\frac{dF}{d{{Q}_{1}}}=0\]Þ \[{{Q}_{1}}={{Q}_{2}}=\frac{Q}{2}\]
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