Question

Two equal charges,distance x apart , exert a force on one another the charge on one of the charges is doubled.what is the ratio of distance between two charges now an earlier if the force remains same in two cases .

Answer 1

Given :

Two equal charges are at a distance of x and they exert force on each other .

The charge on one of the charge  is doubled such that the force remains constant.

To Find:

Ratio of distance between the two charges now to the earlier so that there is no change in the magnitude of force .

Solution :

Let the two charges be $q_{1}$ and $q_{2}$ ;

so $q_{1 } =q$  and also $q_{2} = q$

ow the initial distance between them is x , so the force exerted by them on each other :

$F=\frac{Kq_{1}q_{2} }{x^{2} }$    - ( 1)

Now since one of the charge is doubled :

So $q_{1}' = 2q$

And the new distance between them so  that the force remains constant :

r=x'

Since the force is same so the new expression for the force F is ;

$F= \frac{Kq_{1}'q_{2} }{x'^{2} }$     - (2)

Dividing equations 1 and 2 , we get :

$\frac{F}{F} =\frac{Kq_{1}q_{2} \times x'^{2} }{x^{2} \times Kq_{1}'q_{2} } \\\\1 = \frac{q_{1}x'^{2} }{x^{2} q_{1}' } \\\\\frac{x'^{2} }{x^{2} } = \frac{q_{1}' }{q_{1} } \\\\\\\frac{x'}{x} = \sqrt{\frac{2q_{1} }{q_{1} } } \\\\\frac{x'}{x} = \sqrt{2}$

So the required ratio of distances between the charges now and earlier is √2.