Considering the Coulomb's Law, the electrical force between two charges that are each 2.5×10⁻⁶ C and separated by 3.0 cm from each other is 62.5 N.
From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.
From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
[tex]F=k\frac{Qq}{d^{2} }[/tex]
where:
The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
In this case, you know that two charges are each 2.5×10⁻⁶ C and separated by 3.0 cm (or 0.03 m, being 1 cm= 0.01 m) from each other.
Replacing in the Coulomb's Law, you get:
F=[tex]9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{2.5x10^{-6} Cx 2.5x10^{-6} C}{(0.03m)^{2} }[/tex]
Solving:
F=[tex]9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{6.25x10^{-12} C^{2} }{(0.03m)^{2} }[/tex]
F=[tex]9x10^{9}\frac{Nm^{2} }{C^{2} } 6.94x10^{-9} \frac{ C^{2} }{m^{2} }[/tex]
F= 62.5 N
Finally, the electrical force between two charges that are each 2.5×10⁻⁶ C and separated by 3.0 cm from each other is 62.5 N.
Learn more about Coulomb's Law:
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