Question

Exercise:

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Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. The charge on each is $6.5\times {10}^7$ C? The radii of A and B are negligible compared to the distance of separation.

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Suppose the spheres A and B in Exercise have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Answer 1

$Q_A=6.5\times {10}^{-7}$
$Q_B=6.5\times {10}^{-7}$
Let the uncharged sphere be C.
$Q_C=0$
After contacting with A, the charge gets divided equally between A and C.
So, the new charges on A and C are $6.5\times {10}^{-7}/2$ and $6.5\times {10}^{-7}/2$, respectively.
Now when we brought into contact C and B, the total charge on B and C again gets divided equally between B and C.
So, the new charges on B and C are $\frac{3.25\times {10}^{-7}+6.5\times {10}^{-7}}{2}$ and $\frac{3.25\times {10}^{-7}+6.5\times {10}^{-7}}{2}$, respectively.
So now,
$Q_A^{\prime }=3.25\times {10}^{-7}$
$Q_B^{\prime }=4.875\times {10}^{-7}$
Electrostatic force now is $F=Q_A^{\prime }{Q}_B^{\prime }/4\pi {ϵ}_o{r}^2$
$=5.7\times {10}^{-8}N$