Question

The excess (equal in number) number of electrons that must be placed on each of two small spheres spaced
3 cm apart with force of repulsion between the spheres to be 109N is​

Answer 1

Answer:

The excess (equal in number) number of electrons that must be placed on each of two small spheres spaced

3 cm apart with force of repulsion between the spheres to be 109N is

Answer 2

Given:

Distance between two spheres , r = 3 cm = 0.3 cm

Force of repulsion , F = $10^{-19}$ N

To find:

Number of excess electrons

Solution:

We need to find the number of elections by using the formula q=ne.  

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

$10^{-19}$ = 9 x $10^{9}$ x q² / 0.09

q² = 10^-19 x 0.09 / 9 x 10^9

q^2 = 10^-28 x 10^-2  

q= 10^-15 C.

Substitute it in q=ne to get n,

n = q/e

n = 10^-15/1.6x10^-19

n = 625.