Question

2 equal charges, 27 micro Coulomb each, are separated by 5 cm. Find force between those.

Answer 1

Answer:

The force between charges is  $F= 2.624*10^3N$.

Explanation:

The Coulomb force $F$ between the two charges $q_1$ and $q_2$ separated by distance $d$ is given by the equation

$F = k\dfrac{q_1q_2}{d^2}$

where $k$ is the coulombs constant, and has the value

$k= 9*10^9N\cdot C\cdot m^2$.

Now in our case

$q_1=q_2=27\mu C =27*10^{-6}C$

and

$d= 5cm =0.05m$,

therefore, the Coulomb force between the charges is

$F =( 9*10^9N\cdot C\cdot m^2)*\dfrac{(27*10^{-6}C)(27*10^{-6}C)}{(0.05m)^2}$

$\boxed{ F= 2.624*10^3N}$