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2 years ago

What is the electric force acting between two charges of -0.0050 C and

Physics

1 answer:

love history [14]2 years ago

6 0

The electric force is $-3.6\cdot 10^8 N$ (attractive)

Explanation:

The magnitude of the electric force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following:

$q_1 = -0.0050 C$ (charge 1)

$q_2 = +0.0050 C$ (charge 2)

r = 0.025 m (distance)

Substituting, we find the electric force between the two charges:

$F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N$

And the negative sign means the force is attractive, since the two charges have opposite sign.

Learn more about electric force:

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