Question

Write Coulomb's law in vector form. Two point charges \(q_1 = 10\,\mu\text{C}\) and \(q_2 = 20\,\mu\text{C}\) are located, respectively, in the x-y plane at points \((0, 0)\) and \((3\,\text{cm}, 4\,\text{cm})\). Find (i) the magnitude of the force exerted on \(q_2\) by \(q_1\), and (ii) the angle that the force makes with the x-axis.

Hint:

Use \(F = kq_1q_2/r^2\) with \(r = 5\) cm. The force is repulsive along the line joining the charges, with \(\tan\theta = 4/3\).

Answer 1

Step 1: Coulomb law in vector form:
The force on charge \(q_2\) due to charge \(q_1\) is
\[ \vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \, \hat{r}_{12} \]
Here \(r\) is the distance between the charges and \(\hat{r}_{12}\) is the unit vector pointing from \(q_1\) to \(q_2\). The force is repulsive for like charges and attractive for unlike charges.

Step 2: Find the separation:
\(q_1\) is at \((0, 0)\) and \(q_2\) is at \((3\ \text{cm}, 4\ \text{cm})\). So
\[ r = \sqrt{3^2 + 4^2} = 5\ \text{cm} = 5 \times 10^{-2}\ \text{m} \]

Step 3: Part (i): Magnitude of the force:
\(q_1 = 10 \times 10^{-6}\ \text{C}\) and \(q_2 = 20 \times 10^{-6}\ \text{C}\).
\[ F = \frac{9 \times 10^{9} \times 10 \times 10^{-6} \times 20 \times 10^{-6}}{(5 \times 10^{-2})^2} = \frac{1.8}{2.5 \times 10^{-3}} = 720\ \text{N} \]

Step 4: Part (ii): Direction of the force:
Both charges are positive, so the force on \(q_2\) is repulsive. It points along the line from \(q_1\) to \(q_2\).
\[ \tan\theta = \frac{4}{3} \quad \Rightarrow \quad \theta = 53^{\circ} \]

Final Answer:
(i) The force on \(q_2\) is 720 N. (ii) It makes an angle of about \(53^{\circ}\) with the x-axis. \[ \boxed{F = 720\ \text{N},\quad \theta \approx 53^{\circ}} \]