Answer:
The force is three fourth as large of initial force
Explanation:
Let assume that that charge on each particle be q and the distance which is fixed between them be r.
Then By using Coulombs law the force[tex]F_1[/tex]between them is
[tex]F_1=\dfrac{q^2}{4\pi \epsilon_0 r^2}[/tex]
When half of the charge is transferred from on to other particle then the charge on the particles are 0.5 q and 1.5 q
Let the force between them be [tex]F_2[/tex]
[tex]F_2=\dfrac{0.5 q\times 1.5 q}{4\pi \epsilon_0 r^2}[/tex]
[tex]F_2=\dfrac{3q^2}{4\pi \epsilon_0 4r^2}[/tex]
Then it can can be seen from taking taking ratio that [tex]\dfrac{F_2}{F_1}=\dfrac{3}{4}[/tex]