Question

The force between two point charges is F, which of the following could be done to make the force between them 1/2 F?

• A. Reduce the charge on each point charge to half of its original value and double the distance between the point charges.
• B. Increase the charge on each point charge to double its original value and double the distance between them.
• C. Increase the charge on each point charge to double its original value and move the point charges so that only half the distance is between them.
• D. Increase the charge on only one point charge to twice its value and double the distance between the point charges.

Answer 1

The correct option is D Increase the charge on only one point charge to twice its value and double the distance between the point charges.

By Coulomb's law, $F=\frac{k{q}_1{q}_2}{r^2}$

For A : $F^{\prime }=\frac{k\left(q_1/2\right)\left(q_2/2\right)}{(2r{)}^2}=F/16$

For B : $F^{\prime }=\frac{k\left(2q_1\right)\left(2q_2\right)}{(2r{)}^2}=F$

For C : $F^{\prime }=\frac{k\left(2q_1\right)\left(2q_2\right)}{(r/2{)}^2}=16F$

For D : $F^{\prime }=\frac{k\left(2q_1\right)\left(q_2\right)}{(2r{)}^2}=F/2$

Thus, option D will correct.