Given that,
Initial force, F = 12 N
First initial charge, q₁ = 3C
First new charge, q₁' = 6C
Initial distance, r = 15 cm
New distance, r' = 45 cm
To find,
The new force of attraction.
Solution,
The force between two charges is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]12=k\dfrac{3\times q_2}{15^2}\ ....(1)[/tex]
Let F' is the new force.
[tex]F'=k\dfrac{q_1'q_2'}{r'^2}[/tex]
[tex]F'=\dfrac{k\times 6\times q_2}{(45)^2}\ ...(2)[/tex]
As q₂ is same in this case.
Dividing equation (1) and (2) :
[tex]\dfrac{F}{F'}=\dfrac{k\dfrac{3q_2}{15^2}}{\dfrac{k\times 6\times q_2}{45^2}}\\\\\dfrac{12}{F'}=4.5\\\\F'=\dfrac{12}{4.5}\\\\F'=2.67\ N[/tex]
So, the new force of attraction is 2.67 N.
