Respuesta :
The magnitude of force is F=91.214nN. The force is Repulsive in nature.
Solution:
According to coulombs law, the force between two charges is express as
F=(K*q1*q2) /r^2
constant K = 9*10^9
Hence for a charge q1=6.82nC = 6.82*10^-9,
q2=3.90*10^-9
distance r=1.62m
F = [(9×10^9) ×(6.82×10^-9) × (3.90×10^-9)] /(1.62^2)
F=(239.382×10^-9)/2.6244
F = 91.214×10^-9N
F = 91.214nN
Since the charges are both positive, the force is repulsive
What is Coulomb's law?
Coulomb listed the following characteristics of the electric force for charges in a resting state: Unlike charges attract each other whereas like charges repel one another. As a result, a positive charge pulls a negative charge toward it whereas two negative charges repel one another. The line between the two charges is where the attraction or repulsion occurs.
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Correct question:
Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 6.82 nc, and the other has a charge of 3.90 nc. (a) What is the magnitude (in N) of the electric force that one particle exerts on the other? (b) is the force attractive or repulsive? O attractive O repulsive
