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If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge.

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Answer:

9000 N

Explanation:

Charge, q1 = q2 = 1 C

Distance, r = 1 km = 1000 m

According to the coulomb's law, the force between the two charged particles is given by

[tex]F =\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}}{r^{2}}[/tex]

By substituting the values

[tex]F =\frac{9\times 10^{9}\times 1\times 1}{1000\times 1000}[/tex]

F = 9000 N

Thus, the force of repulsion between the charges is 9000 N.

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