Question

How are electric forces and distance related?

Answer 1

Hint:Coulomb’s law: The magnitude of the electric force between two electric charges ${q_1}$ and ${q_2}$ is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.

Complete step by step answer:
The electric force ${F_E}$, can be either attractive or repulsive, while a gravitational force ${F_g}$, is always attractive. Like charges such as two negative charges or two positive charges will repulse each other. Opposite charges attract each other.
The force of attraction or repulsion depends upon the magnitude of charge and the distance between the two charges
This relation was first studied by coulomb then he gave this relation based on his experiments
$\left| {{F_E}} \right| = k\left| {\dfrac{{{q_1}{q_2}}}{{{r^2}}}} \right|$
Where F
${F_E}$is electric force
K is the Coulomb’s law constant,
${q_1},{q_2}$are the charges, and r is the distance between the charges.
From the above equation, we can draw two conclusions
Electric force directly depends upon the magnitude of the charges i.e., ${F_E} \propto \left| {{q_1}} \right|,\left| {{q_2}} \right|$
The electric force is inversely proportional to the square of the distance between the charges i.e., ${F_E} \propto \left| {\dfrac{1}{{{r^2}}}} \right|$
So we can say that as the distance between the two charges decreases the electric force between them increases parabolically.

Note:
The force increases parabolically because the proportionality is of second-order as the energy is inversely proportional to the square of the distance
K is the coulomb's constant whose value is taken as $k = 9.0 \times {10^9}N.{m^2}{C^{ - 2}}$
We always write the magnitude of charge with its sign convention but in this relationship as we use modulus so we are only concentrating upon their magnitude not on type of charge