Question

If two charges of 1 coulomb each are placed 1 km apart, the force between them will be
(1)$$9\times {10}^3$$N (2)$$9\times {10}^{-3}$$ (3)$$1.1\times {10}^{-4}$$ (4)$${10}^{-6}$$

Answer 1

Hint: We will use the formula of coulomb's electrostatic force between two charges. By substituting the values of k, charge and the distance between them we can find the value of Force between the two charges.
Formula: $F=k{\displaystyle \frac{q_1{q}_2}{r^2}}$

Complete answer:
We know that electrostatic force acts between two stationary electrically charged particles separated by a distance ‘r’. This force is also called coulomb's force. It is given by the formula
$F=k{\displaystyle \frac{q_1{q}_2}{r^2}}$ ---------(1)
Where k is the coulomb's constant, $q_1$and $q_2$is the charge in coulombs.
In the question, it is given that $q_1=q_2=1C$
And the charges are placed 1km apart, so $r=1km=1000m$
The value of coulomb’s constant $k=9\times {10}^{9}N{m}^2{C}^{-2}$
Substituting above values in equation (1), we get
$\begin{array}{rl}& \Rightarrow 9\times {10}^9\times {\displaystyle \frac{1\times 1}{{(1000)}^2}}N\\ & \Rightarrow 9\times {10}^9\times {10}^{-6}N\\ & \Rightarrow 9\times {10}^{3}N\\ & \therefore F=9\times {10}^{3}N\end{array}$
So, If two charges of 1 coulomb are placed 1 km apart, the force between them will be$9\times {10}^{3}N$.

So, Option (1) is correct.

Additional information:
Electrostatic force is an attractive or repulsive force which acts between stationary electrically charged particles. The electric force between a stationary charged body is conventionally known as the electrostatic force or Coulomb’s force.
1 Coulomb charge is equal to the amount of charge from a current of one ampere flowing for one second. It can also be defined as One coulomb is equal to the charge on $6.241\times {10}^{18}$ protons. The charge on 1 proton is $1.6\times {10}^{-19}$ C.

Note:
In the formula for electrostatic force, ‘k’ is the proportionality constant also known as coulomb’s constant.
$$\begin{array}{rl}& k={\displaystyle \frac{1}{4\pi {\xi }_0}}=9\times {10}^{9}N{m}^2{C}^{-2}\\ & {\xi }_0=8.85\times {10}^{-12}{C}^2/N{m}^2\end{array}$$
${\xi }_0$ is known as the permittivity of free space.
One must remember all these values while solving questions of electrostatics.