Question

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of $50\,\text{cm}$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times10^{-7}\,\text{C}$? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Hint:

Use Coulomb's law \(F = kq^2/r^2\). For part (b), doubling \(q\) and halving \(r\) gives \(F' = 4\times4\,F = 16F\).

Answer 1

Given: charge on each sphere \(q = 6.5\times10^{-7}\,\text{C}\); centre-to-centre separation \(r = 50\,\text{cm} = 0.50\,\text{m}\). The radii are negligible, so each sphere acts like a point charge.

Step 1: Coulomb's law. The force of repulsion between two point charges is

\[ F = \frac{k\,q_1 q_2}{r^{2}}, \qquad k = 9\times10^{9}\,\text{N m}^{2}\,\text{C}^{-2}. \]

Step 2 (part a): substitute. Here \(q_1 = q_2 = 6.5\times10^{-7}\,\text{C}\):

\[ F = \frac{(9\times10^{9})(6.5\times10^{-7})(6.5\times10^{-7})}{(0.50)^{2}}. \]

Step 3: compute the numerator and denominator.

\[ (6.5\times10^{-7})^{2} = 4.225\times10^{-13}. \]

\[ \text{numerator} = (9\times10^{9})(4.225\times10^{-13}) = 3.8025\times10^{-3}. \]

\[ \text{denominator} = (0.50)^{2} = 0.25. \]

\[ F = \frac{3.8025\times10^{-3}}{0.25} = 1.52\times10^{-2}\,\text{N}. \]

So the repulsion is about \(1.52\times10^{-2}\,\text{N}\).

Step 4 (part b): double the charge, halve the distance. New charges \(q' = 2q\) and new distance \(r' = r/2\). Then

\[ F' = \frac{k(2q)(2q)}{(r/2)^{2}} = \frac{k\cdot 4q^{2}}{r^{2}/4} = 16\,\frac{kq^{2}}{r^{2}} = 16F. \]

Step 5: compute.

\[ F' = 16\times(1.52\times10^{-2}) = 0.2434\,\text{N} \approx 0.243\,\text{N}. \]

\[\boxed{F = 1.52\times10^{-2}\,\text{N}, \quad F' = 16F \approx 0.243\,\text{N}}\]

Answer 2

Expert method: scaling argument plus direct check for part (b).

Step 1: Because the sphere radii are negligible next to \(0.50\,\text{m}\), each charged sphere is a point charge and Coulomb's law applies exactly: \(F = kq^{2}/r^{2}\).

Step 2 (part a): Insert the numbers once:

\[ F = \frac{(9\times10^{9})(6.5\times10^{-7})^{2}}{(0.50)^{2}} = \frac{3.8025\times10^{-3}}{0.25} = 1.52\times10^{-2}\,\text{N}. \]

Step 3 (part b by scaling): Write \(F \propto q^{2}/r^{2}\). If \(q \to 2q\) the factor from charge is \((2)^{2} = 4\). If \(r \to r/2\) the factor from distance is \((1/(1/2))^{2} = 2^{2} = 4\). Multiplying the two factors gives \(4\times4 = 16\), so \(F' = 16F\).

Step 4 (independent direct check): With \(q' = 1.3\times10^{-6}\,\text{C}\) and \(r' = 0.25\,\text{m}\):

\[ F' = \frac{(9\times10^{9})(1.3\times10^{-6})^{2}}{(0.25)^{2}} = \frac{(9\times10^{9})(1.69\times10^{-12})}{0.0625} = \frac{1.521\times10^{-2}}{0.0625} = 0.2434\,\text{N}. \]

Step 5: The direct value \(0.243\,\text{N}\) equals \(16\times(1.52\times10^{-2}\,\text{N})\), confirming the scaling argument.

\[\boxed{F = 1.52\times10^{-2}\,\text{N}, \quad F' \approx 0.243\,\text{N} = 16F}\]