Question

Two point charges \(q_1 = 3\) $\mu$C and \(q_2 = -4\) $\mu$C are placed at points \((2\hat{i} + 3\hat{j} + 3\hat{k})\) and \((\hat{i} + \hat{j} + \hat{k})\) respectively. Force on charge \(q_2\) is \underline{\hspace{2cm}} N. (Take \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) SI Units)

• A. \((12\hat{i} + 24\hat{j} + 24\hat{k}) \times 10^{-3}\)
• B. \((4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}\)
• C. \((3\hat{i} + 6\hat{j} + 6\hat{k}) \times 10^{-3}\)
• D. \((-4\hat{i} - 8\hat{j} - 8\hat{k}) \times 10^{-3}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to Coulomb's Law in vector form, the force exerted by charge $q_1$ on $q_2$ is directed along the line joining them. Since the charges have opposite signs, the force will be attractive (directed towards $q_1$).
Step 2: Key Formula or Approach:
1. Position vectors: $\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$.
2. Relative vector $\vec{r}_{21} = \vec{r}_1 - \vec{r}_2$.
3. Vector Force $\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{12}$ or use $\vec{F} = k \frac{q_1 q_2}{r^2} \hat{r}$.
Step 3: Detailed Explanation:
$\vec{r}_{21} = (2-1)\hat{i} + (3-1)\hat{j} + (3-1)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.
Magnitude $|\vec{r}_{21}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$ m.
Force on $q_2$ due to $q_1$: \[ \vec{F} = 9 \times 10^9 \frac{(3 \times 10^{-6})(-4 \times 10^{-6})}{3^2} \frac{(\hat{r}_2 - \hat{r}_1)}{|\dots|} \] Actually, for attraction, the force on $q_2$ is towards $q_1$: \[ \vec{F} = \frac{k q_1 |q_2|}{r^3} (\vec{r}_1 - \vec{r}_2) = \frac{9 \times 10^9 \times 3 \times 10^{-6} \times 4 \times 10^{-6}}{3^3} (\hat{i} + 2\hat{j} + 2\hat{k}) \] \[ \vec{F} = \frac{108 \times 10^{-3}}{27} (\hat{i} + 2\hat{j} + 2\hat{k}) = 4 \times 10^{-3} (\hat{i} + 2\hat{j} + 2\hat{k}) \] \[ \vec{F} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3} \text{ N} \]
Step 4: Final Answer:
The force is \((4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}\) N.