Question

If two charges of 1 coulomb each are placed 1 km apart in a vacuum, the force between them will be:
A) $9 \times {10^3}\,N$
B) $9 \times {10^{ - 3}}\,N$
C) $1.1 \times {10^{ - 4}}\,N$
D) ${10^{ - 6}}\,N$

Answer 1

Hint
The force between two charged particles is calculated using the coulomb’s law which quantifies the attraction between two static charged particles. The force is proportional to the product of the charged particles and inversely proportional to the square of the distance between them. So substituting the values from the question we will get the answer.

Formula used: In this solution we will be using the following formula,
Coulomb's law: $F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$ where $F$ is the force acting between two charged particles of charge ${q_1}$ and ${q_2}$ which have a distance $r$ between them.

Complete step by step answer
The force between two static charged particles is calculated using coulomb’s law which is given as
$\Rightarrow F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$
We’ve been given that two charges that have charge 1 coulomb each are placed 1km apart.
Taking the distance in meters since we will be using the SI units, the distance between the two charged particles will be $1\,km\, = 1000m$. Substituting the value of the Coulomb’s constant $k = 9 \times {10^9}$, ${q_1} = {q_2} = 1\,C$ and $r = 1000m$, we get the force as,
$\Rightarrow F = \dfrac{{9 \times {{10}^9} \times 1 \times 1}}{{{{\left( {{{10}^3}} \right)}^2}}}$
On calculating we get,
$\Rightarrow F = \dfrac{{9 \times {{10}^9}}}{{{{10}^6}}}$
Simplifying the force, we get
$F = 9 \times {10^3}$ which corresponds to option (A).

Additional Information
Here, we substituted the value of $k$ for vacuum that exists between the two charges. If there was a medium between the two charges, the correct value of $k$ should be substituted taking into account the permittivity of the medium.

Note
Since we’ve not been given the polarity of the charges i.e. whether they are positive or negative, we can only find out the magnitude of the force between them and not whether the two charges will attract or repel each other. Here, we have also assumed that there are no other charges present between and around the two charges in the vacuum.