Answer:
The charge with greater magnitude = [tex]3.60\times 10^{-5}C[/tex]
Explanation:
We have the distance between two charges = 30 cm = 0.3 m
Force between two charges = 65 N
Let the first charge be q
Then according to question second charge will be 2q
According to coulombs law force between two charges
[tex]F=\frac{Kq_1q_2}{r^2}[/tex]
Where K is constant which value is [tex]K=9\times 10^9[/tex]
So [tex]65=\frac{9\times 10^9\times 2q^2}{0.3^2}[/tex]
[tex]q^2=3.25\times 10^{-10}[/tex]
[tex]q=1.80\times 10^{-5}C[/tex]
So larger charge [tex]2q=2\times 1.80\times 10^{-5}=3.60\times 10^{-5}C[/tex]
