Answer:
F = -10800 N
Explanation:
Given that,
Charge 1, [tex]q_1=-1.5\ C[/tex]
Charge 2, [tex]q_2=0.8\ C[/tex]
Distance between the charges, [tex]d=1\ km=10^3\ m[/tex]
We need to find the electric force acting between two point charges. Mathematically, it is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=-9\times 10^9\times \dfrac{1.5\times 0.8}{(10^3)^2}[/tex]
F = -10800 N
So, the magnitude of electric force between two point charges is 10800 N. Hence, this is the required solution.
