Question

Question: What would be the magnitude of the electrostatic force between two 1.00Cpoint charges separated by a distance of (a) 1 .00 mand (b) 1.00 kmif such point charges existed (they do not) and this configuration could be set up?

Short answer

Answer:

1. The magnitude of the electrostatic force for distance of 1.00 m is$9.00\times {10}^{9}N$.
2. The magnitude of the electrostatic force for distance of 1.00 km is$9.00\times {10}^{3}N$.

Step-by-step solution

Stating given data

Two point charges of charge are separated by the distance.

1. r = 1.00 m
2. $r=1.00km\text{or}1000m$.

Understanding concept of Coulomb’s law

Using the concept of Coulomb's law, we can find the magnitude of the required force between the particles.

Formula:

Magnitude of electrostatic force between any two particles,$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$. (i)

a) Calculation of magnitude of force for 

Using the given data in equation (i), we get the magnitude of the force as follows:

$$\begin{gathered} F=\frac{\left(9\times {10}^{9}N{m}^2/C^2\right){\left(1.00C\right)}^2}{{\left(1.00m\right)}^2} \\ =9.00\times {10}^{9}N \end{gathered}$$

Hence, the value of the force is$9.00\times {10}^{9}N$

b) Calculation of magnitude of force for 

Using the given data in equation (i), we get the magnitude of the force as follows:

$$\begin{gathered} F=\frac{\left(9\times {10}^{9}N{m}^2/C^2\right){\left(1.00C\right)}^2}{{\left(1.00\times {10}^{3}m\right)}^2} \\ =9.00\times {10}^{3}N \end{gathered}$$

Hence, the value of the force is$9.00\times {10}^{3}N$