Answer:
New force = 0.063 N
Explanation:
Given that,
The electric force between two charges is[tex]4.2\times 10^{-2}\ N[/tex]
The formula for the electric force is:
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
If the distance between the charges is doubled, r' = 2r and one of the charges is tripled, q₁' = 2q₁, q₂' = 3q₂
Put all the values,
[tex]F'=\dfrac{kq_1'q_2'}{r'^2}\\\\\dfrac{F}{F'}=\dfrac{\dfrac{kq_1q_2}{r^2}}{\dfrac{kq_1'q_2'}{r'^2}}\\\\\dfrac{F}{F'}=\dfrac{\dfrac{q_1\times q_2}{r^2}}{\dfrac{2q_1\times 3q_2}{(2r)^2}}\\\\\dfrac{F}{F'}=\dfrac{4}{6}=\dfrac{2}{3}\\\\F'=\dfrac{3\times 4.2\times 10^{-2}}{2}\\\\F'=0.063\ N[/tex]
So, the new force is 0.063 N.
Answer:
The force becomes 0.0315 N.
Explanation:
Force, F = 4.2 x 10^-2 N
When the distance is doubled, a charge is tripled, Let the force is F'.
The force between the two charges is
[tex]F = \frac{K qq'}{r^2}\\[/tex]
when, q' = 3 q' and r is 2 r so
[tex]F' = \frac{K 3qq'}{4r^2} = \frac{3 F}{4} = \frac {3\times 4.2\times 10^{-2}}{4}=0.0315 N[/tex]
