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

Answer:

Number of electrons, n = 395.47

Explanation:

It is given that,

Force between two spheres, $F=2.2\times 10^{-21}\ N$

Distance between spheres, r = 35 cm = 0.35 m

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

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

$q^2=\dfrac{F.r^2}{k}$

$q^2=\dfrac{2.2\times 10^{-21}\ N\times (0.35\ m)^2}{9\times 10^9\ Nm^2/C^2}$

$q^2=2.99\times 10^{-32}$

$q=1.72\times 10^{-16}\ C$

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

q = n e

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

$n=\dfrac{1.72\times 10^{-16}}{1.6\times 10^{-19}}$

n = 395.47

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