Question

A small brass sphere having a positive charge of 1.7x10-8C is made to touch another sphere of the same radius having a negative charge of 3x10^-9 C. Find the force between them when they are separated by a distance of 20cm .

Answer 1

see the attachment for the answer

Answer 2

Answer:

The force between them when they are separated by a distance is $11.03\times10^{-6}\ N$.

Explanation:

Given that,

Positive charge $q_{1}=1.7\times10^{-8}\ C$

Negative charge $q_{2}=3\times10^{-9}\ C$

Distance r = 20 cm

After contact, the charges on both the spheres will be same.

The charge will be

$Q=\dfrac{+17\times10^{-9}+(-3\times10^{-9})}{2}$

$Q=7\times10^{-9}$

The force between them

$F= \dfrac{k\times Q^2}{r^2}$

$F=\dfrac{9\times10^{9}\times(7\times10^{-9})^2}{(0.2)^2}$

$F= 11.03\times10^{-6}\ N$

Hence, The force between them when they are separated by a distance is $11.03\times10^{-6}\ N$.