# 6.2 Coulomb’s law  (Page 2/6)

 Page 2 / 6

As the example implies, gravitational force is completely negligible on a small scale, where the interactions of individual charged particles are important. On a large scale, such as between the Earth and a person, the reverse is true. Most objects are nearly electrically neutral, and so attractive and repulsive Coulomb forces nearly cancel. Gravitational force on a large scale dominates interactions between large objects because it is always attractive, while Coulomb forces tend to cancel.

## Section summary

• Frenchman Charles Coulomb was the first to publish the mathematical equation that describes the electrostatic force between two objects.
• Coulomb’s law gives the magnitude of the force between point charges. It is
$F=k\frac{|{q}_{1}{q}_{2}|}{{r}^{2}},$

where ${q}_{1}$ and ${q}_{2}$ are two point charges separated by a distance $r$ , and $k\approx 8.99×{10}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}·{\text{m}}^{2}/{\text{C}}^{2}$

• This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.
• The Coulomb force is extraordinarily strong compared with the gravitational force, another basic force—but unlike gravitational force it can cancel, since it can be either attractive or repulsive.
• The electrostatic force between two subatomic particles is far greater than the gravitational force between the same two particles.

## Conceptual questions

[link] shows the charge distribution in a water molecule, which is called a polar molecule because it has an inherent separation of charge. Given water’s polar character, explain what effect humidity has on removing excess charge from objects.

Using [link] , explain, in terms of Coulomb’s law, why a polar molecule (such as in [link] ) is attracted by both positive and negative charges.

Given the polar character of water molecules, explain how ions in the air form nucleation centers for rain droplets.

## Problems&Exercises

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

(a) How strong is the attractive force between a glass rod with a $0.700\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge and a silk cloth with a $–0.600\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges.

(a) 0.263 N

(b) If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force.

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?

The separation decreased by a factor of 5.

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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
how did you get the value of 2000N.What calculations are needed to arrive at it
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