A charge of 4,mu C is to be divided into two | vedclass

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(B) Let the total charge $Q = 4\,\mu C$ be divided into two parts $q$ and $(Q - q)$.The electrostatic force $F$ between these two charges separated by

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$2\,\mu C$ and $2\,\mu C$

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(B) Let the total charge $Q = 4\,\mu C$ be divided into two parts $q$ and $(Q - q)$.The electrostatic force $F$ between these two charges separated by a distance $d$ is given by Coulomb's law:$F = \frac{K q (Q - q)}{d^2}$To find the condition for maximum force,we differentiate $F$ with respect to $q$ and set it to zero:$\frac{dF}{dq} = \frac{K}{d^2} \frac{d}{dq} (Qq - q^2) = 0$$\frac{K}{d^2} (Q - 2q) = 0$Since $K$ and $d$ are constants and non-zero,we have:$Q - 2q = 0 \implies q = \frac{Q}{2}$Given $Q = 4\,\mu C$,we get $q = \frac{4\,\mu C}{2} = 2\,\mu C$.Thus,the two charges are $2\,\mu C$ and $2\,\mu C$.

$A$ charge of $4\,\mu C$ is to be divided into two parts. The distance between the two divided charges is constant. What should be the magnitude of the divided charges so that the electrostatic force between them is maximum?

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