The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
ured from the planet's surface). What type of relationship exists between the altitude and the atmospheric density, and what would the atmospheric density be at an altitude of 1,291 kilometers?
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Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.
<h3>Coulomb's Law</h3>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:
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
.
The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
<h3>This case</h3>In this case, you know that:
- The two uncharged sphere are separated by the distance of d= 3.50 m
- The number of electrons are 1.30×10¹².
- Electrons is elementary charge and charges on both the sphere is same. The value of electron is 1.602×10⁻¹⁹ C. This is, Q=q=1.30×10¹²×1.602×10⁻¹⁹ C= 2.0826×10⁻⁷ C
Replacing in Coulomb's Law:
Solving:
<u><em>F= 3.1865 N</em></u>
Finally, the magnitude of the Coulomb force is 3.1865 N.
Learn more about Coulomb's Law:
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