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.

Answer 2

The number of excess electrons that should be present is 887.5.

Calculation of no of excess electrons:

Since the force of repulsion between two spheres is $4.57\times10^{-21}$

Now the force of attraction should be$F = kq^2 \div r^2$

here we assume that the separation between the spheres is 20 cm or 0.2 m

So,

$q = \sqrt{Fr^2 \div k}\\\\ = \sqrt{4.57 \times 10^{-21}\times (0.2)^2 \div 9\times 10^9} \\\\= 1.42 \times 10^{-16}C$

Here we assume the number of excess electrons should be n

So,

q = ne

$n= q\div e\\\\= 1.42 \times 10^{-16} \div 1.6 \times 10^{-19}$

= 887.5

Learn more about the force here: https://brainly.com/question/21422036