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Multiple Choice
If the force between two charges is F when the distance is d, what will the force between the two charges be if they were moved to a distance of 2d?
A
2F
B
F/2
C
F/4
D
4F
Verified step by step guidance
1
Start by recalling Coulomb's Law, which states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by \( F = \frac{k \cdot q_1 \cdot q_2}{r^2} \), where \( k \) is Coulomb's constant.
In the initial scenario, the force \( F \) is given when the distance between the charges is \( d \). Thus, \( F = \frac{k \cdot q_1 \cdot q_2}{d^2} \).
Now, consider the new scenario where the distance between the charges is increased to \( 2d \). Substitute \( 2d \) into the formula: \( F' = \frac{k \cdot q_1 \cdot q_2}{(2d)^2} \).
Simplify the expression for \( F' \): \( F' = \frac{k \cdot q_1 \cdot q_2}{4d^2} \).
Compare the new force \( F' \) with the original force \( F \). Since \( F = \frac{k \cdot q_1 \cdot q_2}{d^2} \), it follows that \( F' = \frac{F}{4} \). Thus, the force is reduced to one-fourth of the original force when the distance is doubled.