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Advocard [28]
3 years ago
6

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 charg

e exerts on the other.(b) Is the force attractive or repulsive?
Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

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}=\frac{kq_1q_2}{d^2}

where k=constant =9\times 10^{9}

F_{12}=F_{21}=\frac{9\times 10^9\times 8.01\times 4.50\times 10^{-18}}{(1.87)^2}

F_{12}=F_{21}=92.769\times 10^{-9}\ N

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

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