Question

Two particles are 15 meters apart.

Answer 1

Considering the Coulomb's Law, the combination of particle A: 12.0×10⁻⁴ C and particle B: 2.5×10⁻⁴ C would yield the same Coulomb force of 12 N.

Coulomb's Law

Charged bodies experience a force of attraction or repulsion on approach.

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:

$F=k\frac{Qq}{d^{2} }$

where:

• F is the electrical force of attraction or repulsion. It is measured in Newtons (N).
• Q and q are the values ​​of the two point charges. They are measured in Coulombs (C).
• d is the value of the distance that separates them. It is measured in meters (m).
• K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ $\frac{Nm^{2} }{C^{2} }$.

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

This case

In this case, you know that two particles are 15 meters apart and the resulting Coulomb force is 12 N.

On the other side, you know:

• Particle A: 8.0×10⁻⁴ C, and Particle B: 3.0×10⁻⁴ C
• Particle A: 12.0×10⁻⁴ C, and Particle B: 2.5×10⁻⁴ C
• Particle A: 9.0×10⁻⁴ C, and Particle B: 8.0×10⁻⁴ C
• Particle A: 9.0×10⁻⁴ C, and Particle B: 3.0×10⁻⁴ C

Replacing in the Coulomb's Law, you get:

• $F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(8x10^{-4}C) (3x10^{-4} C )}{(15 m)^{2} }$ →  F= 9.6 N
• $F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(12x10^{-4}C) (2.5x10^{-4} C )}{(15 m)^{2} }$ → F= 12 N
• $F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(9x10^{-4}C) (8x10^{-4} C )}{(15 m)^{2} }$ → F= 28.8 N
• $F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(9x10^{-4}C) (3x10^{-4} C )}{(15 m)^{2} }$ → F=10.8 N

Finally, the combination of particle A: 12.0×10⁻⁴ C and particle B: 2.5×10⁻⁴ C would yield the same Coulomb force of 12 N.

Learn more about Coulomb's Law:

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