Question

two small spheres spaced 35cm apart have equal charge how many excess elecotrons must be present on each sphere if the magnitude of the force of repulsion between them is 2.20x10-21N?

Answer 1

Number of electrons, n = 395.47

Step-by-step explanation:

It is given that,

Force between two spheres,
`F=2.2* 10^(-21)\ N`

Distance between spheres, r = 35 cm = 0.35 m

A force of repulsion is acting on the spheres. It is given by :

`F=k(q^2)/(r^2)`

`q^2=(F.r^2)/(k)`

`q^2=(2.2* 10^(-21)\ N* (0.35\ m)^2)/(9* 10^9\ Nm^2/C^2)`

`q^2=2.99* 10^(-32)`

`q=1.72* 10^(-16)\ C`

Let n is the number of electrons on the spheres. So,

q = n e

`n=(q)/(e)`

`n=(1.72* 10^(-16))/(1.6* 10^(-19))`

n = 395.47

So, the the number of excess electrons on the spheres are 395.47. Hence, this is the required solution.