Suppose the spheres A and B in question 1.12 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?

Solution

Charge on sphere A, q_{A = $6.5 \times 10^{- 7} C$}
Charge on sphere B, q_b= 6.5 x 10^–7 C

When a similar but uncharged sphere C is placed in contact with sphere A, each sphere shares a charge q/2, equally.

charge on sphere A or C = \frac{1}{2} (q + 0) = \frac{q}{2}

Now, if the sphere C is placed in contact with sphere B afterwards, the charge is equally redistributed, so that
Charge on sphere B or C =

\frac{1}{2} (q + q / 2) = 3 q / 4

Thus, the force of repulsion between A and B according to Coulomb's law is (as per fig. above)

F = \frac{1}{4 {πε}_{0}} . \frac{\frac{3 q}{4} . q / 2}{(r / 2)^{2}} = \frac{3}{8} . \frac{1}{4 {πε}_{0}} . \frac{q^{2}}{r^{2}} = \frac{3}{8} \times 1.5 \times 10^{- 2} N = 0.5625 \times 10^{- 2} N = 5.7 \times 10^{- 3} N .

For Daily Free Study Material Join wiredfaculty