Answer:
F' = 64 F
Explanation:
The electric force between charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Where
q₁ and q₂ are charges
r is the distance between charges
When each charge is doubled and the distance between them is 1/4 its original magnitude such that,
q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)
New force,
[tex]F'=\dfrac{kq_1'q_2'}{r'^2}[/tex]
Apply new values,
[tex]F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F[/tex]
So, the new force becomes 64 times the initial force.
