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(B) Let the original charges on spheres $A$ and $B$ be $q_{1}$ and $q_{2}$ respectively.Let the distance between the two spheres be $r$.The initial fo

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greater than that before the balls touched

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(B) Let the original charges on spheres $A$ and $B$ be $q_{1}$ and $q_{2}$ respectively.Let the distance between the two spheres be $r$.The initial force between them is $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$.Since both spheres are identical,when they are brought into contact,the total charge is redistributed equally between them.Thus,the new charge on each sphere is $q^{\prime} = \frac{q_{1} + q_{2}}{2}$.The new force of repulsion between spheres $A$ and $B$ is $F^{\prime} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{\prime} q^{\prime}}{r^{2}} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(\frac{q_{1} + q_{2}}{2})^{2}}{r^{2}}$.According to the Arithmetic Mean-Geometric Mean inequality,for any two positive numbers $q_{1}$ and $q_{2}$ where $q_{1} 
eq q_{2}$,we have $(\frac{q_{1} + q_{2}}{2})^{2} > q_{1} q_{2}$.Therefore,$F^{\prime} > F$.

Two identical conducting balls $A$ and $B$ have positive charges $q_{1}$ and $q_{2}$ respectively,where $q_{1} \neq q_{2}$. The balls are brought together so that they touch each other and then separated to their original positions. The force between them is

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