Answer to Question #197070 in Electric Circuits for harsh

Question

19.Two insulated charged copper spheres A and B have their centres separated

by a distance of 50 cm. What is the mutual force of electrostatic repulsion if

the charge on each is 6.5 × 10–7C.Suppose the spheres A and B 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

Force of repulsion between the two spheres:

$F=\frac{q_Aq_B}{4\pi\varepsilon_0r^2}=\frac{9\cdot10^{9}\cdot(6.5\cdot10^{-7})^2}{0.5^2}=1.52\cdot10^{-8}\ N$

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 $6.5\cdot10^{-7}/2$ and $6.5\cdot10^{-7}/2$

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

$\frac{3.25\cdot10^{-7}+6.5\cdot10^{-7}}{2}$ and $\frac{3.25\cdot10^{-7}+6.5\cdot10^{-7}}{2}$

So now:

$q'_A=3.25\cdot10^{-7}\ C,\ q'_B=4.875\cdot10^{-7}\ C$

Now, the force of repulsion between A and B:

$F=\frac{q'_Aq'_B}{4\pi\varepsilon_0r^2}=\frac{9\cdot10^{9}\cdot3.25\cdot10^{-7}\cdot4.875\cdot10^{-7}}{0.5^2}=5.7\cdot10^{-8}\ N$