Question

$\begin{array}{c}\text{Suppose the spheres}A\text{and}B\text{in above question have identical sizes. A third sphere of the}\\ \text{same size but uncharged is brought in contact with the first, then brought in contact with the}\\ \text{second, and finally removed from both. What is the new force of repulsion between A and B?}\end{array}$

Answer 1

The charge on $A$ and $B,q_1=q_2=6.5\times {10}^{7}C$.

The distance between the sphere $r=0.5m$. Since the spheres are same size, they will possess equal charges on being brought in contact. When uncharged sphere $c(q_3=0)$ is bought in contact with $A$,m charge left on $A$.

$q_1^{\prime }=\frac{q_1+q_3}{2}=\frac{6.5\times {10}^{-7}+0}{2}=3.25\times {10}^{-7}C$

and charge on sphere $C,q_3^1=3.25\times {10}^{-7}C$

Therefore, when the sphere $C$ is brought in contact with $B$, charge left on sphere $B$,

$q_2^{\prime }=\frac{q_2+q_3^{\prime }}{2}=\frac{6.5\times {10}^{-7}+3.25\times {10}^{-7}}{2}$

$=4.875\times {10}^{-7}C$

Therefore, new force of repulsion between the sphere $A$ and $B$,

$F=\frac{1}{4\pi {\in }_0}\times \frac{q_1^1\times q_2^1}{r^2}$

$=9\times {10}^9\times \frac{3.25\times {10}^{-7}\times 4.875\times {10}^{-7}}{(0.5{)}^2}$

$=5.704\times {10}^{-3}N$.