Question

The electrostatic force on a small sphere of charge $0.4\,\mu\text{C}$ due to another small sphere of charge $-0.8\,\mu\text{C}$ in air is $0.2\,\text{N}$.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Hint:

Invert Coulomb's law: \(r=\sqrt{k|q_1q_2|/F}\). For (b) use Newton's third law: the second sphere feels the same magnitude, here 0.2 N, attractive.

Answer 1

We are given the force and the two charges and must find the separation, then use Newton's third law for part (b).

Step 1: Coulomb's law for the magnitude of the force is
\[F=k\,\frac{|q_1 q_2|}{r^2}\quad\Rightarrow\quad r=\sqrt{\frac{k\,|q_1 q_2|}{F}}\]

Step 2 (part a): List the data.
\(q_1=0.4\ \mu\text{C}=0.4\times10^{-6}\ \text{C}\)
\(q_2=-0.8\ \mu\text{C}=0.8\times10^{-6}\ \text{C}\ (\text{magnitude})\)
\(F=0.2\ \text{N}\).

Step 3: Substitute into the rearranged formula.
\[r=\sqrt{\frac{(9\times10^{9})(0.4\times10^{-6})(0.8\times10^{-6})}{0.2}}\]

Step 4: Evaluate the numerator inside the root.
\((9\times10^{9})(0.4\times10^{-6})(0.8\times10^{-6})=9\times0.32\times10^{9-12}=2.88\times10^{-3}\)

Step 5: Divide by \(F\) and take the square root.
\[r=\sqrt{\frac{2.88\times10^{-3}}{0.2}}=\sqrt{1.44\times10^{-2}}=0.12\ \text{m}=12\ \text{cm}\]

Step 6 (part b): By Newton's third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere. Since the charges are unlike, the force is attractive on both.
So the force on the second sphere is \(0.2\ \text{N}\) (attractive).

\[\boxed{r=0.12\ \text{m}=12\ \text{cm},\qquad F_{2\to}=0.2\ \text{N (attractive)}}\]

Answer 2

Expert approach: dimensional ratio for (a), symmetry of mutual forces for (b).

Step 1: Instead of plugging into a square root blindly, isolate \(r^2\) and read it as a ratio of "force available" to "force required".
\[r^2=\frac{k\,|q_1 q_2|}{F}\]
The grouped charge product is \(|q_1 q_2|=(0.4)(0.8)\times10^{-12}=0.32\times10^{-12}\ \text{C}^2\).

Step 2: Compute \(r^2\) keeping powers of ten separate.
\[r^2=\frac{9\times10^{9}\times 0.32\times10^{-12}}{0.2}=\frac{2.88\times10^{-3}}{0.2}=1.44\times10^{-2}\ \text{m}^2\]

Step 3: Recognise \(1.44\times10^{-2}=(0.12)^2\), so \(r=0.12\ \text{m}\). No calculator is needed because \(144\) is a perfect square.

Step 4 (part b): The two spheres form an action-reaction pair. The electrostatic interaction is mutual: each charge feels a force of the same magnitude. Hence \(|\vec F_{21}|=|\vec F_{12}|=0.2\ \text{N}\); the unlike signs make both forces attractive, pointing toward each other. This is a direct consequence of Newton's third law, which Coulomb's law respects automatically.

\[\boxed{r=12\ \text{cm},\qquad F_2=0.2\ \text{N (attractive)}}\]