Question

electric force of repulsion on a tiny charged conducting sphere A, as a function of its speration from a sphere B, the sphere B has 10 times the charge on sphere A, explain the behaviour of force between separation 2 cm and one CM

Answer 1

Force between charges is given by Coloumb's law.
$f \alpha (q1 \times q2) \div {r}^{2}$

where q1 charge of first sphere and q2 charge of second sphere and r is the distance between them.

when r=1cm force is F1
when r=2cm force is F2

q2 = 10 x q1

$\frac{f1}{f2} = \frac{4}{1}$
so f1 will be 4 times that of f2

As distance between charges increases, force decreases by square of the distance between them and vice versa.

k is proportionality constant
F= k*(q1*q2)/r2

F1=k*(q1*10q1)/1
F2=k*(q1*10q1)/4

F1/F2= 4/1