Question

What is the force between two small charged spheres having charges of $2\times10^{-7}\,\text{C}$ and $3\times10^{-7}\,\text{C}$ placed $30\,\text{cm}$ apart in air?

Hint:

Apply \(F=k\,q_1q_2/r^2\) with \(k=9\times10^{9}\), charges in coulomb and \(r=0.30\) m. Like charges give a repulsive force.

Answer 1

We use Coulomb's law for the force between two point charges in air.

Step 1: State the formula. The magnitude of the electrostatic force between two charges \(q_1\) and \(q_2\) separated by a distance \(r\) is
\[F=\frac{1}{4\pi\epsilon_0}\,\frac{q_1 q_2}{r^2}=k\,\frac{q_1 q_2}{r^2}\]
with \(k=9\times10^{9}\ \text{N m}^2\,\text{C}^{-2}\).

Step 2: List the data.
\(q_1=2\times10^{-7}\ \text{C}\)
\(q_2=3\times10^{-7}\ \text{C}\)
\(r=30\ \text{cm}=0.30\ \text{m}\).

Step 3: Substitute the values.
\[F=\frac{(9\times10^{9})(2\times10^{-7})(3\times10^{-7})}{(0.30)^2}\]

Step 4: Evaluate the numerator and denominator.
Numerator: \(9\times10^{9}\times 2\times10^{-7}\times 3\times10^{-7}=54\times10^{-5}=5.4\times10^{-4}\)
Denominator: \((0.30)^2=0.09\)

Step 5: Divide.
\[F=\frac{5.4\times10^{-4}}{0.09}=6\times10^{-3}\ \text{N}\]

Both charges are positive, so the force is repulsive and acts along the line joining them.

\[\boxed{F=6\times10^{-3}\ \text{N (repulsive)}}\]

Answer 2

Expert approach: order-of-magnitude scaling plus direction from the vector form.

Step 1: Write Coulomb's law in vector form so the direction is unambiguous.
\[\vec F_{12}=k\,\frac{q_1 q_2}{r^2}\,\hat r\]
where \(\hat r\) points from charge 1 to charge 2. Since \(q_1 q_2>0\), \(\vec F\) is along \(+\hat r\), i.e. each sphere is pushed away from the other (repulsion).

Step 2: Group the powers of ten before multiplying the mantissas, which keeps the arithmetic clean.
\[F=k\,\frac{q_1 q_2}{r^2}=\big(9\times10^{9}\big)\,\frac{(2)(3)\times10^{-14}}{9\times10^{-2}}\]

Step 3: The factor \(9\) in \(k\) cancels the \(9\times10^{-2}\) in the denominator.
\[F=\frac{9}{9}\times10^{9}\times 6\times10^{-14}\times10^{2}=6\times10^{9-14+2}=6\times10^{-3}\ \text{N}\]

Step 4: Sanity check by magnitude: charges of order \(10^{-7}\ \text{C}\) at tens of centimetres give milli-newton forces, which matches \(6\times10^{-3}\ \text{N}\).

\[\boxed{\vec F=6\times10^{-3}\ \text{N along the line of centres, repulsive}}\]