Step 1 : Understanding the Question:
This problem concerns the electrostatic interaction between two static point charges separated by a certain distance. We are asked to evaluate how the electrostatic force changes when the distance of separation between these charges is doubled.
Step 2 : Key Formulas and Approach:
The electrostatic force between two stationary point charges is governed by Coulomb's Law. This fundamental law states that the force of attraction or repulsion is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The equation is written as:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) represent the magnitudes of the charges, and \( r \) is the separation distance. This shows an inverse-square relationship:
\[ F \propto \frac{1}{r^2} \]
Step 3 : Detailed Explanation:
Let us define the initial state of the system where the distance between the two charges is \( r \). The initial electrostatic force is represented by \( F_1 = \frac{k q_1 q_2}{r^2} \).
According to the problem statement, the distance between the charges is doubled, which gives us a new distance of \( r' = 2r \).
We can write the expression for the new electrostatic force \( F_2 \) by substituting the new distance \( r' \) into the formula: \( F_2 = \frac{k q_1 q_2}{(r')^2} = \frac{k q_1 q_2}{(2r)^2} \).
Simplifying the denominator by squaring the term gives: \( F_2 = \frac{k q_1 q_2}{4r^2} \).
To find the relationship between the two forces, we can factor out the constant from the expression: \( F_2 = \frac{1}{4} \left( \frac{k q_1 q_2}{r^2} \right) \).
Substituting the initial force \( F_1 \) back into this equation yields: \( F_2 = \frac{1}{4} F_1 \).
Step 4 : Final Answer:
The electrostatic force between the two charges becomes one-fourth of the original force, which corresponds to option (A).