Question

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−214.57×10−21 newtons?

Answer 1

Answer:

There are 887.5 number of excess electrons on the each spheres.

Explanation:

It is given that, the force of repulsion between two spheres is, $F=4.57\times 10^{-21}\ N$

We know that the force of attraction or repulsion between charges is given by :

$F=\dfrac{kq^2}{r^2}$

It is assumed that the separation between the spheres is 20 cm or 0.2 m. So,

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{4.57\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$    

$q=1.42\times 10^{-16}\ C$

Let there are n excess electrons that must be present on each sphere. It can be given by using quantization of charge as :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{1.42\times 10^{-16}}{1.6\times 10^{-19}}$

n = 887.5 electrons

So, there are 887.5 number of excess electrons on the each spheres. Hence, this is the required solution.