Question

Suppose the spheres A and B in Exercise 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?

Hint:

Touching identical conductors splits their total charge equally. A becomes q/2; then the third sphere (q/2) shared with B gives B = 3q/4. Plug both into Coulomb's law at the same r = 0.5 m.

Answer 1

From Exercise 1.12, spheres A and B each carry the same charge \(q = 6.5\times10^{-7}\,\text{C}\) and are separated by \(r = 0.5\,\text{m}\). The original force is \(F_0 = 1.52\times10^{-2}\,\text{N}\).

Step 1: When two identical conducting spheres touch, the total charge is shared equally between them. The uncharged third sphere C (charge \(0\)) first touches A.

\[q_A' = \frac{q + 0}{2} = \frac{q}{2} = \frac{6.5\times10^{-7}}{2} = 3.25\times10^{-7}\,\text{C}\]

Sphere C now carries \(q/2\) as well.

Step 2: Sphere C (charge \(q/2\)) is next touched to B (charge \(q\)). They share equally.

\[q_B' = \frac{q + \tfrac{q}{2}}{2} = \frac{3q}{4} = \frac{3 \times 6.5\times10^{-7}}{4} = 4.875\times10^{-7}\,\text{C}\]

Step 3: After C is removed, A holds \(q_A' = 3.25\times10^{-7}\,\text{C}\) and B holds \(q_B' = 4.875\times10^{-7}\,\text{C}\). Apply Coulomb's law with \(k = 9\times10^{9}\):

\[F = \frac{k\,q_A'\,q_B'}{r^{2}}\]\[F = \frac{9\times10^{9} \times 3.25\times10^{-7} \times 4.875\times10^{-7}}{(0.5)^{2}}\]\[F = \frac{9\times10^{9} \times 1.584\times10^{-13}}{0.25}\]\[F = 5.7\times10^{-3}\,\text{N}\]

The repulsion is still mutual and along the line AB.

\[\boxed{F \approx 5.7\times10^{-3}\,\text{N}}\]

Answer 2

Ratio method (no constants needed for the last step):

Step 1: Because both spheres start with the same charge \(q\), write the new charges as simple fractions of \(q\). Touching A with the neutral sphere halves A: \(q_A' = \tfrac{1}{2}q\). The neutral sphere leaves A with \(\tfrac{1}{2}q\).

Step 2: That sphere (\(\tfrac{1}{2}q\)) shares with B (\(q\)): \(q_B' = \tfrac{1}{2}\left(q + \tfrac{1}{2}q\right) = \tfrac{3}{4}q\).

Step 3: Since the separation \(r\) is unchanged, take the ratio of new force to old force \(F_0 = \dfrac{k q^{2}}{r^{2}}\):

\[\frac{F}{F_0} = \frac{k\,q_A' q_B'/r^{2}}{k\,q^{2}/r^{2}} = \frac{\left(\tfrac{1}{2}q\right)\left(\tfrac{3}{4}q\right)}{q^{2}} = \frac{3}{8}\]

Step 4: Multiply by the known original force \(F_0 = 1.52\times10^{-2}\,\text{N}\):

\[F = \frac{3}{8}\,F_0 = 0.375 \times 1.52\times10^{-2} = 5.7\times10^{-3}\,\text{N}\]

This shortcut avoids re-substituting \(k\), since only the charge fractions changed.

\[\boxed{F = \tfrac{3}{8}F_0 \approx 5.7\times10^{-3}\,\text{N}}\]