# The force between two charges 0.06m apart is 5N. If each charge is moved towards the other by 0.04m,then the force between them will become?

Susmita
425 Points
3 years ago
Hey let q and Q be the charge.r=0.06 is the distance between them.
So the force $f=5=\frac{qQ}{4\pi\epsilon r^2}=\frac{qQ}{4\pi\epsilon (0.06)^2}$
Let us draw a straight line.Now draw 7 equidistant point on line:0,0.01,0.02,0.03,0.04,0.05,0.06.The charge q is located at 0 and Q is located at 0.06.
If you move 0.04 unit from point 0 to the right then you will reach at the point 0.04.If you move 0.04 unit from the point 0.06 ,you will reach the point 0.02.
So now the distance between the charges is (0.04-0.02)=0.02.So the force now is
$f'=\frac{qQ}{4\pi\epsilon (0.02)^2}$
So

$\frac{f'}{f} =\frac{(0.06)^2}{ (0.02)^2}$
Or,$f'=f\frac{(0.06)^2}{ (0.02)^2}=5 (\frac{3}{1})^2 =5*9=45N$
Hope this helps