Answer:
F'= 4F/9
Explanation:
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :
[tex]F=\dfrac{kQ^2}{r^2}[/tex] ...(1)
Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :
[tex]F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}[/tex].....(2)
Dividing equation (1) and (2), we get :
[tex]\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}[/tex]
Hence, the correct option is (d) i.e. " 4F/9"
