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sineoko [7]
3 years ago
10

A 7.94 nC charge is located 1.79 m from a 4.06 nC point charge. (a) Find the magnitude of the electrostatic force that one charg

e exerts on the other. N (b) Is the force attractive or repulsive? attractive repulsive
Physics
1 answer:
mixer [17]3 years ago
6 0

Answer:

9.044824943\times 10^{-8}\ N

repulsive

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q_1 = 7.94 nC

q_2 = 4.06 nC

r = Distance between the particles = 1.79 m

Electrical force force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 7.94\times 10^{-9}\times 4.06\times 10^{-9}}{1.79^2}\\\Rightarrow F=9.044824943\times 10^{-8}\ N

The magnitude of the electrostatic force is 9.044824943\times 10^{-8}\ N

Both the paricles have positive charges this means they will repel each other.

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3 years ago
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