Answer:
The force between charges is [tex]F= 2.624*10^3N[/tex].
Explanation:
The Coulomb force [tex]F[/tex] between the two charges [tex]q_1[/tex] and [tex]q_2[/tex] separated by distance [tex]d[/tex] is given by the equation
[tex]F = k\dfrac{q_1q_2}{d^2}[/tex]
where [tex]k[/tex] is the coulombs constant, and has the value
[tex]k= 9*10^9N\cdot C\cdot m^2[/tex].
Now in our case
[tex]q_1=q_2=27\mu C =27*10^{-6}C[/tex]
and
[tex]d= 5cm =0.05m[/tex],
therefore, the Coulomb force between the charges is
[tex]F =( 9*10^9N\cdot C\cdot m^2)*\dfrac{(27*10^{-6}C)(27*10^{-6}C)}{(0.05m)^2}[/tex]
[tex]\boxed{ F= 2.624*10^3N}[/tex]
