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  • State Coulomb’s law in terms of how the electrostatic force changes with the distance between two objects.
  • Calculate the electrostatic force between two charged point forces, such as electrons or protons.
  • Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and the Earth.
Two spiral galaxies show the strong gravitational attraction between them as their arms appear to reach out toward one another.
This NASA image of Arp 87 shows the result of a strong gravitational attraction between two galaxies. In contrast, at the subatomic level, the electrostatic attraction between two objects, such as an electron and a proton, is far greater than their mutual attraction due to gravity. (credit: NASA/HST)

Through the work of scientists in the late 18th century, the main features of the electrostatic force    —the existence of two types of charge, the observation that like charges repel, unlike charges attract, and the decrease of force with distance—were eventually refined, and expressed as a mathematical formula. The mathematical formula for the electrostatic force is called Coulomb’s law    after the French physicist Charles Coulomb (1736–1806), who performed experiments and first proposed a formula to calculate it.

Coulomb’s law

F = k | q 1 q 2 | r 2 . size 12{F=k { {q rSub { size 8{1} } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {}

Coulomb’s law calculates the magnitude of the force F between two point charges, q 1 size 12{q rSub { size 8{1} } } {} and q 2 size 12{q rSub { size 8{2} } } {} , separated by a distance r . In SI units, the constant k is equal to

k = 8 . 988 × 10 9 N m 2 C 2 8 . 99 × 10 9 N m 2 C 2 . size 12{k=8 "." "988" times "10" rSup { size 8{9} } { {N cdot m rSup { size 8{2} } } over {C rSup { size 8{2} } } } approx 9 "." "00" times "10" rSup { size 8{9} } { {N cdot m rSup { size 8{2} } } over {C rSup { size 8{2} } } } } {}

The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges. (See [link] .)

Although the formula for Coulomb’s law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment then available, were difficult. Modern experiments have verified Coulomb’s law to great precision. For example, it has been shown that the force is inversely proportional to distance between two objects squared F 1 / r 2 size 12{ left (F prop {1} slash {r rSup { size 8{2} } } right )} {} to an accuracy of 1 part in 10 16 size 12{"10" rSup { size 8{"16"} } } {} . No exceptions have ever been found, even at the small distances within the atom.

In part a, two charges q one and q two are shown at a distance r. Force vector arrow F one two points toward left and acts on q one. Force vector arrow F two one points toward right and acts on q two. Both forces act in opposite directions and are represented by arrows of same length. In part b, two charges q one and q two are shown at a distance r. Force vector arrow F one two points toward right and acts on q one. Force vector arrow F two one points toward left and acts on q two. Both forces act toward each other and are represented by arrows of same length.
The magnitude of the electrostatic force F size 12{F} {} between point charges q 1 size 12{q rSub { size 8{1} } } {} and q 2 size 12{q rSub { size 8{2} } } {} separated by a distance r size 12{F} {} is given by Coulomb’s law. Note that Newton’s third law (every force exerted creates an equal and opposite force) applies as usual—the force on q 1 size 12{q rSub { size 8{1} } } {} is equal in magnitude and opposite in direction to the force it exerts on q 2 size 12{q rSub { size 8{2} } } {} . (a) Like charges. (b) Unlike charges.

How strong is the coulomb force relative to the gravitational force?

Compare the electrostatic force between an electron and proton separated by 0 . 530 × 10 10 m size 12{0 "." "530" times "10" rSup { size 8{ - "10"} } m} {} with the gravitational force between them. This distance is their average separation in a hydrogen atom.

Strategy

To compare the two forces, we first compute the electrostatic force using Coulomb’s law, F = k | q 1 q 2 | r 2 size 12{F=k { {q rSub { size 8{1} } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {} . We then calculate the gravitational force using Newton’s universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude.

Solution

Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb’s law yields

F = k | q 1 q 2 | r 2 size 12{F=k { {q rSub { size 8{1 } } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {}
= 8.99 × 10 9 N m 2 / C 2 × ( 1.60 × 10 –19 C ) ( 1.60 × 10 –19 C ) ( 0.530 × 10 –10 m ) 2 alignl { stack { size 12{" "= left (9 "." "00 " times " 10" rSup { size 8{9} } N cdot " m" rSup { size 8{2} } /C rSup { size 8{2} } right ) times { { \( "-1" "." "60 " times " 10" rSup { size 8{"-19"} } C \) \( 1 "." "60" times " 10" rSup { size 8{"-19 "} } C \) } over { \( 0 "." "530 " times " 10" rSup { size 8{"-10"} } m \) rSup { size 8{2} } } } } {} #{} } } {}

Thus the Coulomb force is

F = 8.19 × 10 –8 N . size 12{F=" -8" "." "20 " times " 10" rSup { size 8{"-8"} } N} {}

The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron—it would cause an acceleration of 8.99 × 10 22 m / s 2 size 12{9 "." "00" times "10" rSup { size 8{"22"} } {m} slash {s rSup { size 8{2} } } } {} (verification is left as an end-of-section problem).The gravitational force is given by Newton’s law of gravitation as:

F G = G mM r 2 , size 12{F rSub { size 8{G} } =" G " { {"mM"} over {r rSup { size 8{2} } } } } {}

where G = 6.67 × 10 11 N m 2 / kg 2 size 12{G=6 "." "67" times "10" rSup { size 8{ - "11"} } {N cdot m rSup { size 8{2} } } slash { ital "kg" rSup { size 8{2} } } } {} . Here m and M represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields

F G = ( 6.67 × 10 11 N m 2 / kg 2 ) × ( 9.11 × 10 –31 kg ) ( 1.67 × 10 –27 kg ) ( 0.530 × 10 –10 m ) 2 = 3.61 × 10 –47 N

This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of the magnitude of the electrostatic force to gravitational force in this case is, thus,

F F G = 2 . 27 × 10 39 . size 12{ { {F} over {F rSub { size 8{G} } } } =" 2" "." "27 " times " 10" rSup { size 8{"39"} } } {}

Discussion

This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature.

Questions & Answers

how do they get the third part x = (32)5/4
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can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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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.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Physics 105: adventures in physics. OpenStax CNX. Dec 02, 2015 Download for free at http://legacy.cnx.org/content/col11916/1.1
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