A certain charge 2Q is divided at first into | vedclass

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(A) Let the total charge be $2Q$. The two parts are $q_{1}$ and $q_{2}$,such that $q_{1} + q_{2} = 2Q$. Thus,$q_{2} = 2Q - q_{1}$.According to Coulomb

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(A) Let the total charge be $2Q$. The two parts are $q_{1}$ and $q_{2}$,such that $q_{1} + q_{2} = 2Q$. Thus,$q_{2} = 2Q - q_{1}$.According to Coulomb's law,the force of interaction $F$ between the charges separated by a distance $r$ is given by $F = k \frac{q_{1}q_{2}}{r^2}$.Substituting $q_{2}$,we get $F = \frac{k}{r^2} (q_{1})(2Q - q_{1}) = \frac{k}{r^2} (2Qq_{1} - q_{1}^2)$.For the force $F$ to be maximum,the derivative of $F$ with respect to $q_{1}$ must be zero: $\frac{dF}{dq_{1}} = 0$.$\frac{d}{dq_{1}} [\frac{k}{r^2} (2Qq_{1} - q_{1}^2)] = \frac{k}{r^2} (2Q - 2q_{1}) = 0$.This implies $2Q - 2q_{1} = 0$,so $q_{1} = Q$.Therefore,the ratio $\frac{Q}{q_{1}} = \frac{Q}{Q} = 1$.

$A$ certain charge $2Q$ is divided at first into two parts $q_{1}$ and $q_{2}$. Later,the charges are placed at a certain distance. If the force of interaction between the two charges is maximum,then find the value of $\frac{Q}{q_{1}}$.

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