Question

Two small charged spheres are 6.52 cm apart. they are moved, and the force each exerts on the other is found to have tripled. how far apart are they now?

Answer 1

So all that really matters is the radius and how it changes, which is by a fact or of √3 and ∴ 6.52/√3 = 3.76cm which is the answer

Answer 2

Answer:      

They are now 3.75 cm apart.

Explanation:

It is given that, two small charged spheres are 6.52 cm apart. It they are moved, the force is found to be become tripled.

The electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

Let initial force is F' so,

$F'=k\dfrac{q_1q_2}{(0.0652)^2}$...........(1)

When they moved apart, let final force is F'' = 3 F'.

$3F'=k\dfrac{q_1q_2}{r'^2}$...................(2)

Taking ratios of equation (1) and (2) :

$r'^2=\dfrac{(0.0652)^2}{3}$

$r'=0.0014\ m$

or $r'=0.0375\ m$

r' = 3.75 cm

Hence, this is the required solution.