Answer:
[tex]\frac{1}{2}F[/tex]
Explanation:
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=9\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem, at the beginning we have:
[tex]q_1=q_A[/tex] is the first charge
[tex]q_2=q_B[/tex] is the second charge
r is their initial separation
So the initial force is
[tex]F=k\frac{q_A q_B}{r^2}[/tex]
Later, we have:
qA is doubled and r is doubled
This means that:
[tex]q_A'=2q_A[/tex] is the new charge
[tex]r'=2r[/tex] is the new separation
So the new force is:
[tex]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[/tex]
