Topic: Electromagnetics
A charge $Q_{2}=8.854\times10^{-9}\text{C}$ is located in vacuum at $P_{2}(2,3,1)$. The force on $Q_{2}$ due to a charge $Q_{1}=4\pi\times10^{-3}\text{C}$ at $P_{1}(2,2,1)$ is
(Note: All the coordinates are measured in metres. $a_{x}$, $a_{y}$ and $a_{z}$ are unit vectors in X, Y and Z direction respectively.)
- A. $a_{y}\text{N}$
- B. $-a_{y}\text{N}$
- C. $4a_{x}+5a_{y}+2a_{z}\text{N}$
- D. $-4a_{x}-5a_{y}-2a_{z}\text{N}$
Answer: A. $a_{y}\text{N}$
Explanation: The distance vector is $\vec{R} = P_2 - P_1 = (2-2)\hat{a}_x + (3-2)\hat{a}_y + (1-1)\hat{a}_z = \hat{a}_y$.
Magnitude $|\vec{R}| = 1\text{ m}$.
The force is $\vec{F} = \frac{Q_1 Q_2}{4\pi\epsilon_0 R²} \hat{a}_y = \frac{4\pi\times 10^{-3} \times 8.854\times 10^{-9}}{4\pi (8.854\times 10^{-12}) \times 1^2} \hat{a}_y = 1 \hat{a}_y \text{ N}$.