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Question

Since the two spheres are identical, final charge on each of them after contact $q=\frac{\left(q_{1}+q_{2}\right)}{2}$

$F_{1}=\frac{K q_{1} q_{2}}{

Vedclass · Accepted Answer

more than before

Vedclass · Answer

Since the two spheres are identical, final charge on each of them after contact $q=\frac{\left(q_{1}+q_{2}\right)}{2}$

$F_{1}=\frac{K q_{1} q_{2}}{r^{2}}$

$F_{2}=\frac{K q^{2}}{r^{2}}=\frac{K}{r^{2}}\left(\frac{q_{1}+q_{2}}{2}\right)^{2}$

$F_{2}-F_{1}=\frac{K}{r^{2}}\left\{\left(\frac{q_{1}+q_{2}}{2}\right)^{2}-q_{1} q_{2}\right\}$

$=\frac{K}{4 r^{2}}\left(q_{1}-q_{2}\right)^{2} \Rightarrow F_{2}-F_{1}>0$ or $F_{2}>F_{1}$

Two identical conducting spheres having unequal positive charges $q_1$ and $q_2$ separated by distance $r$. If they are made to touch each other and then separated again to the same distance, the electrostatic force between them in this case will be :-

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