Answer:
Fe = 7.98*10^5N/C
Explanation:
In order to calculate the electric force between the two charges you use the following formula:
[tex]F_e=k\frac{q_1q_2}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q1 = 2mC = 2*10^-3C
q2 = 4mC = 4*10^-3C
r: distance between the charge = 30cm = 0.3m
You replace the values of the parameters in the equation (1):
[tex]F_e=(8.98*10^9Nm^2/C^2)\frac{(2*10^{-3}C)(4*10^{-3}C)}{(0.3)^2}\\\\F_e=7.98*10^5\frac{N}{C}[/tex]
The force between the charges q1 and q2 is 7.98*10^5N/C
