Question

Electrostatic force of two charge particles depends on
A. the separation between the charges
B. the magnitude of the charges
C. mass of the charges
D. both A and B

Answer 1

Hint: The electric charge of any system is an integral multiple of the amount of the charge. They can neither be created nor destroyed. The force between the charges is explained by Coulomb’s law of electrostatic force.

Complete step-by-step solution -
Electrostatic force is also called coulomb force, deals with static electric charges or charges at rest. Coulomb’s law states that the force of attraction or repulsion between two-point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The directions of forces are along the line joining the two charges.
Consider $q_1$ and $q_2$ as two charges placed with a distance ‘r’ between them. Then, according to Coulomb’s law,
$F\alpha \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$ or
$F=k\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$
Now, let us substitute the value of ‘k’ in the above-mentioned formula. We get,
$F=\dfrac { 1 }{ 4\pi { \in }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$
Where, $\dfrac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{c}^{-2}}$ . On solving the equation, with $q_1$ and $q_2$ as 1C and r =1m
$F=\left( 9\times {{10}^{9}} \right)\dfrac{1}{{{1}^{2}}}=9\times {{10}^{9}}N$
One coulomb is defined as the quantity of charge, which when placed at a distance of 1 meter from another charge will experience a repulsive force of $9\times {{10}^{9}}N$ .
Additional Information: If the charges are present in a medium of permittivity , the magnitude of force between the charges will be,
${{F}_{m}}=\dfrac{1}{4\pi \in }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ where $F_m$ is the force between the charges in a medium.
On dividing both the equations, we get
$\dfrac{F}{{{F}_{m}}}=\dfrac{\in }{{{\in }_{0}}}={{\in }_{r}}$ . This ratio is called the relative permittivity of the dielectric constant of the medium.
So, the correct answer to the given question is option (D).

Note: Since ${{F}_{m}}=\dfrac{F}{{{\in }_{r}}}$, the force between the charges totally depend on the nature of the medium in which they are placed. Based on the electrostatic behaviour, materials are divided into two categories as conductors and insulators.