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(B) Let the total charge $q = 10\,\mu C$ be divided into two parts $x$ and $(q - x)$.The electrostatic repulsive force between them at a distance $r$

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$5\,\mu C , 5\,\mu C$

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(B) Let the total charge $q = 10\,\mu C$ be divided into two parts $x$ and $(q - x)$.The electrostatic repulsive force between them at a distance $r$ is given by Coulomb's Law:$F = \frac{K x(q - x)}{r^2}$To find the condition for maximum force,we differentiate $F$ with respect to $x$ and set it to zero:$\frac{dF}{dx} = \frac{K}{r^2} \frac{d}{dx} (qx - x^2) = 0$$\frac{K}{r^2} (q - 2x) = 0$Since $\frac{K}{r^2} 
eq 0$,we have $q - 2x = 0$,which implies $x = \frac{q}{2}$.Given $q = 10\,\mu C$,the two parts are:$x = \frac{10\,\mu C}{2} = 5\,\mu C$$(q - x) = 10\,\mu C - 5\,\mu C = 5\,\mu C$.Thus,the charges are $5\,\mu C$ and $5\,\mu C$.

$A$ $10\,\mu C$ charge is divided into two parts and placed at $1\,cm$ distance so that the repulsive force between them is maximum. The charges of the two parts are :

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