Question

Two charges, qA and qB, are separated by a distance, r, and exert a force, F, on each other. What new force will exist for the following scenario? Your answer should be given in multiples of F (i.e. 2F, ½F, etc.)
qA is doubled and r is doubled

Answer 1

Answer:

$\frac{1}{2}F$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, at the beginning we have:

$q_1=q_A$ is the first  charge

$q_2=q_B$ is the second charge

r is their initial separation

So the initial force is

$F=k\frac{q_A q_B}{r^2}$

Later, we have:

qA is doubled and r is doubled

This means that:

$q_A'=2q_A$ is the new charge

$r'=2r$ is the new separation

So the new force is:

$F'=k\frac{q_A' q_B}{r'^2}=k\frac{(2q_A)(q_B)}{(2r)^2}=\frac{1}{2}(k\frac{q_A q_B}{r^2})=\frac{1}{2}F$