A charge Q is divided into two charges q and | vedclass

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(C) According to Coulomb's law,the electrostatic force $F$ between two charges $q$ and $Q-q$ separated by a distance $r$ is given by:$F = \frac{1}{4 \

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$\frac{Q}{2}$

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(C) According to Coulomb's law,the electrostatic force $F$ between two charges $q$ and $Q-q$ separated by a distance $r$ is given by:$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q(Q-q)}{r^2}$To find the value of $q$ for which the force is maximum,we differentiate $F$ with respect to $q$ and set it to zero:$\frac{dF}{dq} = \frac{1}{4 \pi \varepsilon_0 r^2} \cdot \frac{d}{dq}(Qq - q^2) = 0$Since $\frac{1}{4 \pi \varepsilon_0 r^2} 
eq 0$,we have:$\frac{d}{dq}(Qq - q^2) = 0$$Q - 2q = 0$$2q = Q$$q = \frac{Q}{2}$Thus,the force is maximum when the charge $Q$ is divided into two equal parts,$q = \frac{Q}{2}$.

$A$ charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum,is

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