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A 8.01-nC charge is located 1.87 m from a 4.50-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other.(b) Is the force attractive or repulsive?

User Dot
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1 Answer

6 votes

Answer:

Step-by-step explanation:

Given


q_1=8.01\ nC


q_2=4.50\ nC

distance between the charges is given by
d=1.87\ m

Electrostatic force between the charges is given by


F_(12)=F_(21)=(kq_1q_2)/(d^2)

where k=constant
=9* 10^(9)


F_(12)=F_(21)=(9* 10^9* 8.01* 4.50* 10^(-18))/((1.87)^2)


F_(12)=F_(21)=92.769* 10^(-9)\ N

as both the charges are of similar nature therefore they repel each other

User Tbondwilkinson
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