Question

Charge q is divided in two parts q₁ and q₂ and these are placed at a distance r. For maximum repulsion between these parts

Easy

Solution

Here one charge is q₁then $q_{2} = (q - q_{1})$
Force between these two charges is given by $F = \frac{1}{4 {πε}_{0}} \times \frac{q_{1} (q - q_{1})}{r^{2}}$
For maximum repulsion $\frac{dF}{{dq}_{1}} = 0 = \frac{1}{4 {πε}_{0}} \times \frac{(q - 2 q_{1})}{r^{2}}$
or $q - 2 q_{1} = 0 or q_{1} = q / 2$ and $q_{2} = q / 2$ .

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