Question

Position vectors of two-point charges of 1nC each are (1, 1, -1) m and (2, 3, 1) m respectively. The magnitude of electric force acting between them is ____N.
\[\begin{align}
  & \text{A) 1}{{\text{0}}^{-3}} \\
 & \text{B) 1}{{\text{0}}^{-6}} \\
 & \text{C) 1}{{\text{0}}^{-9}} \\
 & \text{D) 1}{{\text{0}}^{-12}} \\
\end{align}\]

Answer 1

Hint: We need to understand the relation between the position of the two charges, their magnitude of charge and the electric force that can exist between them in order to solve this given problem easily. The Coulomb’s force law can be used directly.

Complete Step-by-Step Solution:
We are given two charges which are placed in specific coordinates in the space. We need to find the electric force acting between them to solve this problem.

We know that Coulomb's law of electrostatic force can give the answer to this directly. According to the law, the force experienced by each of the charged particles is proportional to the product of the magnitude of their charges and inversely proportional to the square of the distance between them.
i.e.,
\[\begin{align}
  & F\propto \dfrac{qQ}{{{r}^{2}}} \\
 & \therefore F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{qQ}{{{r}^{2}}} \\
\end{align}\]

We can find the distance between the two given charges using the coordinates applying the distance formula to them in three dimensions as –
\[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}\]

We are given the coordinates of the charges as – (1, 1, -1) m and (2, 3, 1) m. We can find the distance between them as –
\[\begin{align}
  & d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}} \\
 & \Rightarrow d=\sqrt{{{(1-2)}^{2}}+{{(1-3)}^{2}}+{{(-1-1)}^{2}}} \\
 & \Rightarrow d=\sqrt{1+4+4} \\
 & \therefore d=3m \\
\end{align}\]

So, we understand that the two charges are 3 m apart from each other. We can calculate the electric force using this information as –
\[\begin{align}
  & F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{qQ}{{{r}^{2}}} \\
 & \Rightarrow F=(9\times {{10}^{9}})\dfrac{1\times {{10}^{-9}}\times 1\times {{10}^{-9}}}{{{3}^{2}}} \\
 & \Rightarrow F=9\times {{10}^{-9}}\times \dfrac{1}{9} \\
 & \therefore F={{10}^{-9}}N \\
\end{align}\]

We can see that the force experienced by the two charges are in the order of \[{{10}^{-9}}\]N.
This is the required solution.

The correct answer is option C.

Note:
We can understand that the electric force is independent of the path taken by the two charges or their initial conditions when finding the force at a particular distance. The shortest distance between the charges will give the required force between them.