Question

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

Answer 1

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

Step-by-step 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(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)*((27*10^(-6)C)(27*10^(-6)C))/((0.05m)^2)`

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