Question

A charge of 1.0C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between two charges to be 2.0km. Find the force exerted by the charges on each other,
A.$2.25\times {10}^{3}N$
B.$2.5\times {10}^{3}N$
C.$225\times {10}^{3}N$
D.$2.25\times {10}^{5}N$

Answer 1

Hint: Force exerted by the charges on each other can be calculated using Coulomb's Law. This law states that the magnitude of electrostatic force of attraction or repulsion between two charges is proportional to the product of charges and inversely proportional to square of distance between them. Substitute the values in the formula and obtain the force exerted by the charges on each other.
Formula used:
$F={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{q_1{q}_2}{r^2}}$

Complete answer:
Given: Separation between two charges (r)= 2 km= 2000m
            Charge $q_1=q_2=1C$
Force of attraction between two charges is given by,
$F={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{q_1{q}_2}{r^2}}$
Substituting the values in above equation we get,
$F={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{1\times 1}{{(2000)}^2}}$
$\Rightarrow F=9\times {10}^9\times {\displaystyle \frac{1}{{4\times 10}^6}}$
$\Rightarrow F={\displaystyle \frac{9}{4}}\times {\displaystyle \frac{{10}^9}{{10}^6}}$
$\Rightarrow F=2.25\times {10}^3$
Thus, the force exerted by the charges on each other is $2.25\times {10}^{3}N$

Hence, the correct answer is option A i.e. $2.25\times {10}^{3}N.$

Note:
The force of attraction between two charges depends upon the distance between both the charges as well the mass of both the particles. It is directly proportional to mass of the bodies. As the separation between the charges or the bodies decreases, the force of attraction between the bodies increases. When the separation between the charges or the bodies increases, the force of attraction between the bodies decreases.