What happens to the force between two charges when you increase the charge on one of the two charges?

Fundamental physical law of electromagnetism

What happens to the force between two charges when you increase the charge on one of the two charges?

The magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Like charges repel each other, and opposite charges mutually attract.

Coulomb's inverse-square law, or simply Coulomb's law, is an experimental law[1] of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force.[2] Although the law was known earlier, it was first published in 1785 by French physicist Charles-Augustin de Coulomb, hence the name. Coulomb's law was essential to the development of the theory of electromagnetism, maybe even its starting point,[1] as it made it possible to discuss the quantity of electric charge in a meaningful way.[3]

The law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them,[4]

| F | = k e | q 1 | | q 2 | r 2 {\displaystyle |F|=k_{\text{e}}{\frac {|q_{1}||q_{2}|}{r^{2}}}}

Here, ke or K is the Coulomb constant (ke ≈ 8.988×109 N⋅m2⋅C−2),[1] q1 and q2 are the signed magnitudes of the charges, and the scalar r is the distance between the charges.

The force is along the straight line joining the two charges. If the charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.

Being an inverse-square law, the law is analogous to Isaac Newton's inverse-square law of universal gravitation, but gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive.[2] Coulomb's law can be used to derive Gauss's law, and vice versa. In the case of a single stationary point charge, the two laws are equivalent, expressing the same physical law in different ways.[5] The law has been tested extensively, and observations have upheld the law on the scale from 10−16 m to 108 m.[5]

History

What happens to the force between two charges when you increase the charge on one of the two charges?

Charles-Augustin de Coulomb

Ancient cultures around the Mediterranean knew that certain objects, such as rods of amber, could be rubbed with cat's fur to attract light objects like feathers and papers. Thales of Miletus made the first recorded description of static electricity around 600 BC,[6] when he noticed that friction could render a piece of amber magnetic.[7][8]

In 1600, English scientist William Gilbert made a careful study of electricity and magnetism, distinguishing the lodestone effect from static electricity produced by rubbing amber.[7] He coined the New Latin word electricus ("of amber" or "like amber", from ἤλεκτρον [elektron], the Greek word for "amber") to refer to the property of attracting small objects after being rubbed.[9] This association gave rise to the English words "electric" and "electricity", which made their first appearance in print in Thomas Browne's Pseudodoxia Epidemica of 1646.[10]

Early investigators of the 18th century who suspected that the electrical force diminished with distance as the force of gravity did (i.e., as the inverse square of the distance) included Daniel Bernoulli[11] and Alessandro Volta, both of whom measured the force between plates of a capacitor, and Franz Aepinus who supposed the inverse-square law in 1758.[12]

Based on experiments with electrically charged spheres, Joseph Priestley of England was among the first to propose that electrical force followed an inverse-square law, similar to Newton's law of universal gravitation. However, he did not generalize or elaborate on this.[13] In 1767, he conjectured that the force between charges varied as the inverse square of the distance.[14][15]

What happens to the force between two charges when you increase the charge on one of the two charges?

Coulomb's torsion balance

In 1769, Scottish physicist John Robison announced that, according to his measurements, the force of repulsion between two spheres with charges of the same sign varied as x−2.06.[16]

In the early 1770s, the dependence of the force between charged bodies upon both distance and charge had already been discovered, but not published, by Henry Cavendish of England.[17]

Finally, in 1785, the French physicist Charles-Augustin de Coulomb published his first three reports of electricity and magnetism where he stated his law. This publication was essential to the development of the theory of electromagnetism.[4] He used a torsion balance to study the repulsion and attraction forces of charged particles, and determined that the magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The torsion balance consists of a bar suspended from its middle by a thin fiber. The fiber acts as a very weak torsion spring. In Coulomb's experiment, the torsion balance was an insulating rod with a metal-coated ball attached to one end, suspended by a silk thread. The ball was charged with a known charge of static electricity, and a second charged ball of the same polarity was brought near it. The two charged balls repelled one another, twisting the fiber through a certain angle, which could be read from a scale on the instrument. By knowing how much force it took to twist the fiber through a given angle, Coulomb was able to calculate the force between the balls and derive his inverse-square proportionality law.

Scalar form of the law

Coulomb's law can be stated as a simple mathematical expression. The scalar form gives the magnitude of the vector of the electrostatic force F between two point charges q1 and q2, but not its direction. If r is the distance between the charges, the magnitude of the force is

| F | = k e | q 1 q 2 | r 2 {\displaystyle |\mathbf {F} |=k_{\text{e}}{\frac {|q_{1}q_{2}|}{r^{2}}}}

The constant ke is called the Coulomb constant and is equal to 1/4πε0, where ε0 is the electric constant; ke ≈ 8.988×109 N⋅m2⋅C−2. If the product q1q2 is positive, the force between the two charges is repulsive; if the product is negative, the force between them is attractive.[18]

Vector form of the law

What happens to the force between two charges when you increase the charge on one of the two charges?

In the image, the vector F1 is the force experienced by q1, and the vector F2 is the force experienced by q2. When q1q2 > 0 the forces are repulsive (as in the image) and when q1q2 < 0 the forces are attractive (opposite to the image). The magnitude of the forces will always be equal.

Coulomb's law in vector form states that the electrostatic force F 1 {\textstyle \mathbf {F} _{1}}

What happens to the force between two charges when you increase the charge on one of the two charges?
experienced by a charge, q 1 {\displaystyle q_{1}}
What happens to the force between two charges when you increase the charge on one of the two charges?
at position r 1 {\displaystyle \mathbf {r} _{1}}
What happens to the force between two charges when you increase the charge on one of the two charges?
, in the vicinity of another charge, q 2 {\displaystyle q_{2}}
What happens to the force between two charges when you increase the charge on one of the two charges?
at position r 2 {\displaystyle \mathbf {r} _{2}}
What happens to the force between two charges when you increase the charge on one of the two charges?
, in a vacuum is equal to[19]

F 1 = q 1 q 2 4 π ε 0 r 1 − r 2 | r 1 − r 2 | 3 = q 1 q 2 4 π ε 0 r ^ 12 | r 12 | 2 {\displaystyle \mathbf {F} _{1}={\frac {q_{1}q_{2}}{4\pi \varepsilon _{0}}}{\frac {\mathbf {r} _{1}-\mathbf {r} _{2}}{|\mathbf {r} _{1}-\mathbf {r} _{2}|^{3}}}={\frac {q_{1}q_{2}}{4\pi \varepsilon _{0}}}{\frac {\mathbf {\hat {r}} _{12}}{|\mathbf {r} _{12}|^{2}}}}

where r 12 = r 1 − r 2 {\textstyle {\boldsymbol {r}}_{12}={\boldsymbol {r}}_{1}-{\boldsymbol {r}}_{2}}

What happens to the force between two charges when you increase the charge on one of the two charges?
is the vectorial distance between the charges, r ^ 12 = r 12 | r 12 | {\textstyle {\widehat {\mathbf {r} }}_{12}={\frac {\mathbf {r} _{12}}{|\mathbf {r} _{12}|}}}
What happens to the force between two charges when you increase the charge on one of the two charges?
a unit vector pointing from q 2 {\textstyle q_{2}}
What happens to the force between two charges when you increase the charge on one of the two charges?
to q 1 {\textstyle q_{1}}
What happens to the force between two charges when you increase the charge on one of the two charges?
,
and ε 0 {\displaystyle \varepsilon _{0}}
What happens to the force between two charges when you increase the charge on one of the two charges?
the electric constant. Here, r ^ 12 {\textstyle \mathbf {\hat {r}} _{12}}
What happens to the force between two charges when you increase the charge on one of the two charges?
is used for the vector notation.

The vector form of Coulomb's law is simply the scalar definition of the law with the direction given by the unit vector, r ^ 12 {\textstyle {\widehat {\mathbf {r} }}_{12}}

What happens to the force between two charges when you increase the charge on one of the two charges?
, parallel with the line from charge q 2 {\displaystyle q_{2}} to charge q 1 {\displaystyle q_{1}} .[20] If both charges have the same sign (like charges) then the product q 1 q 2 {\displaystyle q_{1}q_{2}}
What happens to the force between two charges when you increase the charge on one of the two charges?
is positive and the direction of the force on q 1 {\displaystyle q_{1}} is given by r ^ 12 {\textstyle {\widehat {\mathbf {r} }}_{12}} ; the charges repel each other. If the charges have opposite signs then the product q 1 q 2 {\displaystyle q_{1}q_{2}} is negative and the direction of the force on q 1 {\displaystyle q_{1}} is − r ^ 12 {\textstyle -{\hat {\mathbf {r} }}_{12}}
What happens to the force between two charges when you increase the charge on one of the two charges?
;
the charges attract each other.

The electrostatic force F 2 {\textstyle \mathbf {F} _{2}}

What happens to the force between two charges when you increase the charge on one of the two charges?
experienced by q 2 {\displaystyle q_{2}} , according to Newton's third law, is F 2 = − F 1 {\textstyle \mathbf {F} _{2}=-\mathbf {F} _{1}}
What happens to the force between two charges when you increase the charge on one of the two charges?
.

System of discrete charges

The law of superposition allows Coulomb's law to be extended to include any number of point charges. The force acting on a point charge due to a system of point charges is simply the vector addition of the individual forces acting alone on that point charge due to each one of the charges. The resulting force vector is parallel to the electric field vector at that point, with that point charge removed.

Force F {\textstyle \mathbf {F} }

What happens to the force between two charges when you increase the charge on one of the two charges?
on a small charge q {\displaystyle q}
What happens to the force between two charges when you increase the charge on one of the two charges?
at position r {\textstyle \mathbf {r} }
What happens to the force between two charges when you increase the charge on one of the two charges?
, due to a system of N {\textstyle N}
What happens to the force between two charges when you increase the charge on one of the two charges?
discrete charges in vacuum is[19]

F ( r ) = q 4 π ε 0 ∑ i = 1 N q i r − r i | r − r i | 3 = q 4 π ε 0 ∑ i = 1 N q i R ^ i | R i | 2 , {\displaystyle \mathbf {F} (\mathbf {r} )={q \over 4\pi \varepsilon _{0}}\sum _{i=1}^{N}q_{i}{\frac {\mathbf {r} -\mathbf {r} _{i}}{|\mathbf {r} -\mathbf {r} _{i}|^{3}}}={q \over 4\pi \varepsilon _{0}}\sum _{i=1}^{N}q_{i}{{\hat {\mathbf {R} }}_{i} \over |\mathbf {R} _{i}|^{2}},}

where q i {\displaystyle q_{i}}

What happens to the force between two charges when you increase the charge on one of the two charges?
and r i {\textstyle \mathbf {r} _{i}}
What happens to the force between two charges when you increase the charge on one of the two charges?
are the magnitude and position respectively of the ith charge, R ^ i {\textstyle {\hat {\mathbf {R} }}_{i}}
What happens to the force between two charges when you increase the charge on one of the two charges?
is a unit vector in the direction of R i = r − r i {\textstyle \mathbf {R} _{i}=\mathbf {r} -\mathbf {r} _{i}}
What happens to the force between two charges when you increase the charge on one of the two charges?
, a vector pointing from charges q i {\displaystyle q_{i}} to q {\displaystyle q} .[20]

Continuous charge distribution

In this case, the principle of linear superposition is also used. For a continuous charge distribution, an integral over the region containing the charge is equivalent to an infinite summation, treating each infinitesimal element of space as a point charge d q {\displaystyle dq}

What happens to the force between two charges when you increase the charge on one of the two charges?
. The distribution of charge is usually linear, surface or volumetric.

For a linear charge distribution (a good approximation for charge in a wire) where λ ( r ′ ) {\displaystyle \lambda (\mathbf {r} ')}

What happens to the force between two charges when you increase the charge on one of the two charges?
gives the charge per unit length at position r ′ {\displaystyle \mathbf {r} '}
What happens to the force between two charges when you increase the charge on one of the two charges?
, and d ℓ ′ {\displaystyle d\ell '}
What happens to the force between two charges when you increase the charge on one of the two charges?
is an infinitesimal element of length,[21]

d q ′ = λ ( r ′ ) d ℓ ′ . {\displaystyle dq'=\lambda (\mathbf {r'} )\,d\ell '.}

For a surface charge distribution (a good approximation for charge on a plate in a parallel plate capacitor) where σ ( r ′ ) {\displaystyle \sigma (\mathbf {r} ')}

What happens to the force between two charges when you increase the charge on one of the two charges?
gives the charge per unit area at position r ′ {\displaystyle \mathbf {r} '} , and d A ′ {\displaystyle dA'}
What happens to the force between two charges when you increase the charge on one of the two charges?
is an infinitesimal element of area,

d q ′ = σ ( r ′ ) d A ′ . {\displaystyle dq'=\sigma (\mathbf {r'} )\,dA'.}

For a volume charge distribution (such as charge within a bulk metal) where ρ ( r ′ ) {\displaystyle \rho (\mathbf {r} ')}

What happens to the force between two charges when you increase the charge on one of the two charges?
gives the charge per unit volume at position r ′ {\displaystyle \mathbf {r} '} , and d V ′ {\displaystyle dV'}
What happens to the force between two charges when you increase the charge on one of the two charges?
is an infinitesimal element of volume,[20]

d q ′ = ρ ( r ′ ) d V ′ . {\displaystyle dq'=\rho ({\boldsymbol {r'}})\,dV'.}

The force on a small test charge q {\displaystyle q} at position r {\displaystyle {\boldsymbol {r}}}

What happens to the force between two charges when you increase the charge on one of the two charges?
in vacuum is given by the integral over the distribution of charge

F ( r ) = q 4 π ε 0 ∫ d q ′ r − r ′ | r − r ′ | 3 . {\displaystyle \mathbf {F} (\mathbf {r} )={\frac {q}{4\pi \varepsilon _{0}}}\int dq'{\frac {\mathbf {r} -\mathbf {r'} }{|\mathbf {r} -\mathbf {r'} |^{3}}}.}

where it must be noted that the "continuous charge" version of Coulomb's law is never supposed to be applied to locations for which | r − r ′ | = 0 {\displaystyle |\mathbf {r} -\mathbf {r'} |=0}

What happens to the force between two charges when you increase the charge on one of the two charges?
because that location would directly overlap with the location of a charged particle (e.g. electron or proton) which is not a valid location to analyze the electric field or potential classically. Charge is always discrete in reality, and the "continuous charge" assumption is just an approximation that is not supposed to allow | r − r ′ | = 0 {\displaystyle |\mathbf {r} -\mathbf {r'} |=0} to be analyzed.

Coulomb constant

The Coulomb constant is a proportionality factor that appears in Coulomb's law as well as in other electric-related formulas. Denoted k e {\displaystyle k_{\text{e}}}

What happens to the force between two charges when you increase the charge on one of the two charges?
, it is also called the electric force constant or electrostatic constant[22] hence the subscript e {\displaystyle e}
What happens to the force between two charges when you increase the charge on one of the two charges?
. When the electromagnetic theory is expressed in the International System of Units, force is measured in newtons, charge in coulombs and distance in meters. The Coulomb constant is given by k e = 1 4 π ε 0 {\textstyle k_{\text{e}}={\frac {1}{4\pi \varepsilon _{0}}}}
What happens to the force between two charges when you increase the charge on one of the two charges?
. The constant ε 0 {\displaystyle \varepsilon _{0}} is the vacuum electric permittivity (also known as "electric constant")[23] in C 2 ⋅ m − 2 ⋅ N − 1 {\displaystyle \mathrm {C^{2}\cdot m^{-2}\cdot N^{-1}} }
What happens to the force between two charges when you increase the charge on one of the two charges?
. It should not be confused with ε r {\displaystyle \varepsilon _{r}}
What happens to the force between two charges when you increase the charge on one of the two charges?
, which is the dimensionless relative permittivity of the material in which the charges are immersed, or with their product ε a = ε 0 ε r {\displaystyle \varepsilon _{a}=\varepsilon _{0}\varepsilon _{r}}
What happens to the force between two charges when you increase the charge on one of the two charges?
, which is called "absolute permittivity of the material" and is still used in electrical engineering.

Prior to the 2019 redefinition of the SI base units, the Coulomb constant was considered to have an exact value:

k e = 1 4 π ε 0 = c 0 2 μ 0 4 π = c 0 2 × 10 − 7   H m − 1 = 8.987 551 787 368 176 4 × 10 9   N m 2 C − 2 . {\displaystyle {\begin{aligned}k_{\text{e}}&={\frac {1}{4\pi \varepsilon _{0}}}={\frac {c_{0}^{2}\mu _{0}}{4\pi }}=c_{0}^{2}\times 10^{-7}\ \mathrm {H\,m^{-1}} \\&=8.987\,551\,787\,368\,176\,4\times 10^{9}\ \mathrm {N\,m^{2}\,C^{-2}} .\end{aligned}}}

Since the 2019 redefinition,[24][25] the Coulomb constant is no longer exactly defined and is subject to the measurement error in the fine structure constant. As calculated from CODATA 2018 recommended values, the Coulomb constant is[26]

k e = 8.987 551 792 3 ( 14 ) × 10 9   N m 2 C − 2 . {\displaystyle k_{\text{e}}=8.987\,551\,792\,3\,(14)\times 10^{9}\ \mathrm {N\,m^{2}\,C^{-2}} .}

In Gaussian units and Lorentz–Heaviside units, which are both CGS unit systems, the constant has different, dimensionless values.

In electrostatic units or Gaussian units the unit charge (esu or statcoulomb) is defined in such a way that the Coulomb constant disappears, as it has the value of one and becomes dimensionless.

k e (Gaussian units) = 1 {\displaystyle k_{\text{e (Gaussian units)}}=1}

In Lorentz–Heaviside units, also called rationalized units, the Coulomb constant is dimensionless and is equal to

k e (Lorentz–Heaviside units) = 1 4 π {\displaystyle k_{\text{e (Lorentz–Heaviside units)}}={\frac {1}{4\pi }}}

Gaussian units are more amenable for microscopic problems such as the electrodynamics of individual electrically charged particles.[27] SI units are more convenient for practical, large-scale phenomena, such as engineering applications.[27]

Limitations

There are three conditions to be fulfilled for the validity of Coulomb's inverse square law:[28]

  1. The charges must have a spherically symmetric distribution (e.g. be point charges, or a charged metal sphere).
  2. The charges must not overlap (e.g. they must be distinct point charges).
  3. The charges must be stationary with respect to each other.

The last of these is known as the electrostatic approximation. When movement takes place, Einstein's theory of relativity must be taken into consideration, and a result, an extra factor is introduced, which alters the force produced on the two objects. This extra part of the force is called the magnetic force, and is described by magnetic fields. For slow movement, the magnetic force is minimal and Coulomb's law can still be considered approximately correct, but when the charges are moving more quickly in relation to each other, the full electrodynamics rules (incorporating the magnetic force) must be considered.

Electric field

What happens to the force between two charges when you increase the charge on one of the two charges?

If two charges have the same sign, the electrostatic force between them is repulsive; if they have different sign, the force between them is attractive.

An electric field is a vector field that associates to each point in space the Coulomb force experienced by a unit test charge.[19] The strength and direction of the Coulomb force F {\textstyle \mathbf {F} } on a charge q t {\textstyle q_{t}}

What happens to the force between two charges when you increase the charge on one of the two charges?
depends on the electric field E {\textstyle \mathbf {E} }
What happens to the force between two charges when you increase the charge on one of the two charges?
established by other charges that it finds itself in, such that F = q t E {\textstyle \mathbf {F} =q_{t}\mathbf {E} }
What happens to the force between two charges when you increase the charge on one of the two charges?
. In the simplest case, the field is considered to be generated solely by a single source point charge. More generally, the field can be generated by a distribution of charges who contribute to the overall by the principle of superposition.

If the field is generated by a positive source point charge q {\textstyle q}

What happens to the force between two charges when you increase the charge on one of the two charges?
, the direction of the electric field points along lines directed radially outwards from it, i.e. in the direction that a positive point test charge q t {\textstyle q_{t}} would move if placed in the field. For a negative point source charge, the direction is radially inwards.

The magnitude of the electric field E can be derived from Coulomb's law. By choosing one of the point charges to be the source, and the other to be the test charge, it follows from Coulomb's law that the magnitude of the electric field E created by a single source point charge Q at a certain distance from it r in vacuum is given by

| E | = k e | q | r 2 {\displaystyle |\mathbf {E} |=k_{\text{e}}{\frac {|q|}{r^{2}}}}

A system N of charges q i {\displaystyle q_{i}} stationed at r i {\textstyle \mathbf {r} _{i}} produces an electric field whose magnitude and direction is, by superposition

E ( r ) = 1 4 π ε 0 ∑ i = 1 N q i r − r i | r − r i | 3 {\displaystyle \mathbf {E} (\mathbf {r} )={1 \over 4\pi \varepsilon _{0}}\sum _{i=1}^{N}q_{i}{\frac {\mathbf {r} -\mathbf {r} _{i}}{|\mathbf {r} -\mathbf {r} _{i}|^{3}}}}

Atomic forces

Coulomb's law holds even within atoms, correctly describing the force between the positively charged atomic nucleus and each of the negatively charged electrons. This simple law also correctly accounts for the forces that bind atoms together to form molecules and for the forces that bind atoms and molecules together to form solids and liquids. Generally, as the distance between ions increases, the force of attraction, and binding energy, approach zero and ionic bonding is less favorable. As the magnitude of opposing charges increases, energy increases and ionic bonding is more favorable.

Relation to Gauss's law

Deriving Gauss's law from Coulomb's law

Strictly speaking, Gauss's law cannot be derived from Coulomb's law alone, since Coulomb's law gives the electric field due to an individual point charge only. However, Gauss's law can be proven from Coulomb's law if it is assumed, in addition, that the electric field obeys the superposition principle. The superposition principle says that the resulting field is the vector sum of fields generated by each particle (or the integral, if the charges are distributed smoothly in space).

Outline of proof

Coulomb's law states that the electric field due to a stationary point charge is:

E ( r ) = q 4 π ε 0 e r r 2 {\displaystyle \mathbf {E} (\mathbf {r} )={\frac {q}{4\pi \varepsilon _{0}}}{\frac {\mathbf {e} _{r}}{r^{2}}}}

where

  • er is the radial unit vector,
  • r is the radius, |r|,
  • ε0 is the electric constant,
  • q is the charge of the particle, which is assumed to be located at the origin.

Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give

E ( r ) = 1 4 π ε 0 ∫ ρ ( s ) ( r − s ) | r − s | 3 d 3 s {\displaystyle \mathbf {E} (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int {\frac {\rho (\mathbf {s} )(\mathbf {r} -\mathbf {s} )}{|\mathbf {r} -\mathbf {s} |^{3}}}\,d^{3}\mathbf {s} }

where ρ is the charge density. If we take the divergence of both sides of this equation with respect to r, and use the known theorem[29]

∇ ⋅ r | r | 3 = − ∇ 2 1 | r | = 4 π δ ( r ) {\displaystyle \nabla \cdot {\frac {\mathbf {r} }{|\mathbf {r} |^{3}}}=-\nabla ^{2}{\frac {1}{|\mathbf {r} |}}=4\pi \delta (\mathbf {r} )}

where δ(r) is the Dirac delta function, the result is

∇ ⋅ E ( r ) = 1 ε 0 ∫ ρ ( s ) δ ( r − s ) d 3 s {\displaystyle \nabla \cdot \mathbf {E} (\mathbf {r} )={\frac {1}{\varepsilon _{0}}}\int \rho (\mathbf {s} )\,\delta (\mathbf {r} -\mathbf {s} )\,d^{3}\mathbf {s} }

Using the "sifting property" of the Dirac delta function, we arrive at

∇ ⋅ E ( r ) = ρ ( r ) ε 0 , {\displaystyle \nabla \cdot \mathbf {E} (\mathbf {r} )={\frac {\rho (\mathbf {r} )}{\varepsilon _{0}}},}

which is the differential form of Gauss' law, as desired.

Note that since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more general than Coulomb's law.

Deriving Coulomb's law from Gauss's law

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).

Outline of proof

Taking S in the integral form of Gauss' law to be a spherical surface of radius r, centered at the point charge Q, we have

∮ S E ⋅ d A = Q ε 0 {\displaystyle \oint _{S}\mathbf {E} \cdot d\mathbf {A} ={\frac {Q}{\varepsilon _{0}}}}

By the assumption of spherical symmetry, the integrand is a constant which can be taken out of the integral. The result is

4 π r 2 r ^ ⋅ E ( r ) = Q ε 0 {\displaystyle 4\pi r^{2}{\hat {\mathbf {r} }}\cdot \mathbf {E} (\mathbf {r} )={\frac {Q}{\varepsilon _{0}}}}

where is a unit vector pointing radially away from the charge. Again by spherical symmetry, E points in the radial direction, and so we get

E ( r ) = Q 4 π ε 0 r ^ r 2 {\displaystyle \mathbf {E} (\mathbf {r} )={\frac {Q}{4\pi \varepsilon _{0}}}{\frac {\hat {\mathbf {r} }}{r^{2}}}}

which is essentially equivalent to Coulomb's law. Thus the inverse-square law dependence of the electric field in Coulomb's law follows from Gauss' law.

In Relativity

Coulomb's law can be used to gain insight of the form of magnetic field generated by moving charges since by special relativity, in certain cases the magnetic field can be shown to be a transformation of forces caused by the electric field. When no acceleration is involved in a particle's history, it can be viewed to obey Coulomb's law on any test particle in its own inertial frame. Coulomb's law can be expanded to moving test particle to be of the same form, which shows discrepancy in Newton's third law as followed but it is not strictly obeyed in the framework of special relativity (yet without violating relativistic-energy conservation).[30] This assumption can be justified by obtaining the correct form of field equations, that is with respect to agreement with maxwell's equations. Considering the charge to be invariant of observer, the electric and magnetic fields of a uniformly moving point charge can hence be derived by the Lorentz transformation of the four force on test charge given by Coulomb's law in the charge's frame of reference and attributing magnetic and electric fields by their definition given by the form of Lorentz force.[31] The fields hence found for uniformly moving point charges are given by:[32]

E = q 4 π ϵ 0 r 3 1 − β 2 ( 1 − β 2 sin 2 ⁡ θ ) 3 / 2 r {\displaystyle \mathbf {E} ={\frac {q}{4\pi \epsilon _{0}r^{3}}}{\frac {1-\beta ^{2}}{(1-\beta ^{2}\sin ^{2}\theta )^{3/2}}}\mathbf {r} }

B = q 4 π ϵ 0 r 3 1 − β 2 ( 1 − β 2 sin 2 ⁡ θ ) 3 / 2 v × r c 2 = v × E c 2 {\displaystyle \mathbf {B} ={\frac {q}{4\pi \epsilon _{0}r^{3}}}{\frac {1-\beta ^{2}}{(1-\beta ^{2}\sin ^{2}\theta )^{3/2}}}{\frac {\mathbf {v} \times \mathbf {r} }{c^{2}}}={\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}}}

where q {\displaystyle q} is the charge of the point source, r {\displaystyle \mathbf {r} }
What happens to the force between two charges when you increase the charge on one of the two charges?
is the position vector from the point source to the point in space, v {\displaystyle \mathbf {v} }
What happens to the force between two charges when you increase the charge on one of the two charges?
is the velocity vector of the charged particle, β {\displaystyle \beta }
What happens to the force between two charges when you increase the charge on one of the two charges?
is the ratio of speed of the charged particle divided by the speed of light and θ {\displaystyle \theta }
What happens to the force between two charges when you increase the charge on one of the two charges?
is the angle between r {\displaystyle \mathbf {r} } and v {\displaystyle \mathbf {v} } .

Note that the expression for electric field reduces to Coulomb's law for non-relativistic speeds of the point charge and that the magnetic field in non-relativistic limit (approximating β ≪ 1 {\displaystyle \beta \ll 1}

What happens to the force between two charges when you increase the charge on one of the two charges?
) can be applied to electric currents to get the Biot–Savart law. These solutions, when expressed in retarded time also correspond to the general solution of maxwell's equations given by solutions of Liénard–Wiechert potential, due to the validity of Coulomb's law within its specific range of application. Also note that the spherical symmetry for gauss law on stationary charges is not valid for moving charges owing to the breaking of symmetry by the specification of direction of velocity in the problem. Maxwell's equations can also be manually verified for the above two equations.[33]

Coulomb potential

Quantum field theory

What happens to the force between two charges when you increase the charge on one of the two charges?

The most basic Feynman diagram for QED interaction between two fermions

The Coulomb potential admits continuum states (with E > 0), describing electron-proton scattering, as well as discrete bound states, representing the hydrogen atom.[34] It can also be derived within the non-relativistic limit between two charged particles, as follows:

Under Born approximation, in non-relativistic quantum mechanics, the scattering amplitude A ( | p ⟩ → | p ′ ⟩ ) {\textstyle {\mathcal {A}}(|\mathbf {p} \rangle \to |\mathbf {p} '\rangle )}

What happens to the force between two charges when you increase the charge on one of the two charges?
is:

A ( | p ⟩ → | p ′ ⟩ ) − 1 = 2 π δ ( E p − E p ′ ) ( − i ) ∫ d 3 r V ( r ) e − i ( p − p ′ ) r {\displaystyle {\mathcal {A}}(|\mathbf {p} \rangle \to |\mathbf {p} '\rangle )-1=2\pi \delta (E_{p}-E_{p'})(-i)\int d^{3}\mathbf {r} \,V(\mathbf {r} )e^{-i(\mathbf {p} -\mathbf {p} ')\mathbf {r} }}

This is to be compared to the:

∫ d 3 k ( 2 π ) 3 e i k r 0 ⟨ p ′ , k | S | p , k ⟩ {\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}e^{ikr_{0}}\langle p',k|S|p,k\rangle }

where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential.

Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with m 0 ≫ | p | {\displaystyle m_{0}\gg |\mathbf {p} |}

What happens to the force between two charges when you increase the charge on one of the two charges?

⟨ p ′ , k | S | p , k ⟩ | c o n n = − i e 2 | p − p ′ | 2 − i ε ( 2 m ) 2 δ ( E p , k − E p ′ , k ) ( 2 π ) 4 δ ( p − p ′ ) {\displaystyle \langle p',k|S|p,k\rangle |_{conn}=-i{\frac {e^{2}}{|\mathbf {p} -\mathbf {p} '|^{2}-i\varepsilon }}(2m)^{2}\delta (E_{p,k}-E_{p',k})(2\pi )^{4}\delta (\mathbf {p} -\mathbf {p} ')}

Comparing with the QM scattering, we have to discard the ( 2 m ) 2 {\displaystyle (2m)^{2}}

What happens to the force between two charges when you increase the charge on one of the two charges?
as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:

∫ V ( r ) e − i ( p − p ′ ) r d 3 r = e 2 | p − p ′ | 2 − i ε {\displaystyle \int V(\mathbf {r} )e^{-i(\mathbf {p} -\mathbf {p} ')\mathbf {r} }d^{3}\mathbf {r} ={\frac {e^{2}}{|\mathbf {p} -\mathbf {p} '|^{2}-i\varepsilon }}}

where Fourier transforming both sides, solving the integral and taking ε → 0 {\displaystyle \varepsilon \to 0}
What happens to the force between two charges when you increase the charge on one of the two charges?
at the end will yield

V ( r ) = e 2 4 π r {\displaystyle V(r)={\frac {e^{2}}{4\pi r}}}

as the Coulomb potential.[35]

However, the equivalent results of the classical Born derivations for the Coulomb problem are thought to be strictly accidental.[36][37]

The Coulomb potential, and its derivation, can be seen as a special case of the Yukawa potential, which is the case where the exchanged boson – the photon – has no rest mass.[34]

Simple experiment to verify Coulomb's law

What happens to the force between two charges when you increase the charge on one of the two charges?

Experiment to verify Coulomb's law.

It is possible to verify Coulomb's law with a simple experiment. Consider two small spheres of mass m {\displaystyle m}

What happens to the force between two charges when you increase the charge on one of the two charges?
and same-sign charge q {\displaystyle q} , hanging from two ropes of negligible mass of length l {\displaystyle l}
What happens to the force between two charges when you increase the charge on one of the two charges?
. The forces acting on each sphere are three: the weight m g {\displaystyle mg}
What happens to the force between two charges when you increase the charge on one of the two charges?
, the rope tension T {\displaystyle \mathbf {T} }
What happens to the force between two charges when you increase the charge on one of the two charges?
and the electric force F {\displaystyle \mathbf {F} }
What happens to the force between two charges when you increase the charge on one of the two charges?
. In the equilibrium state:

T sin ⁡ θ 1 = F 1 {\displaystyle \mathbf {T} \sin \theta _{1}=\mathbf {F} _{1}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(1)

and

T cos ⁡ θ 1 = m g {\displaystyle \mathbf {T} \cos \theta _{1}=mg}

What happens to the force between two charges when you increase the charge on one of the two charges?

(2)

Dividing (1) by (2):

sin ⁡ θ 1 cos ⁡ θ 1 = F 1 m g ⇒ F 1 = m g tan ⁡ θ 1 {\displaystyle {\frac {\sin \theta _{1}}{\cos \theta _{1}}}={\frac {F_{1}}{mg}}\Rightarrow F_{1}=mg\tan \theta _{1}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(3)

Let L 1 {\displaystyle \mathbf {L} _{1}}

What happens to the force between two charges when you increase the charge on one of the two charges?
be the distance between the charged spheres; the repulsion force between them F 1 {\displaystyle \mathbf {F} _{1}}
What happens to the force between two charges when you increase the charge on one of the two charges?
, assuming Coulomb's law is correct, is equal to

F 1 = q 2 4 π ε 0 L 1 2 {\displaystyle F_{1}={\frac {q^{2}}{4\pi \varepsilon _{0}L_{1}^{2}}}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(Coulomb's law)

so:

q 2 4 π ε 0 L 1 2 = m g tan ⁡ θ 1 {\displaystyle {\frac {q^{2}}{4\pi \varepsilon _{0}L_{1}^{2}}}=mg\tan \theta _{1}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(4)

If we now discharge one of the spheres, and we put it in contact with the charged sphere, each one of them acquires a charge q 2 {\textstyle {\frac {q}{2}}}

What happens to the force between two charges when you increase the charge on one of the two charges?
. In the equilibrium state, the distance between the charges will be L 2 < L 1 {\textstyle \mathbf {L} _{2}<\mathbf {L} _{1}}
What happens to the force between two charges when you increase the charge on one of the two charges?
and the repulsion force between them will be:

F 2 = ( q 2 ) 2 4 π ε 0 L 2 2 = q 2 4 4 π ε 0 L 2 2 {\displaystyle F_{2}={\frac {{({\frac {q}{2}})}^{2}}{4\pi \varepsilon _{0}L_{2}^{2}}}={\frac {\frac {q^{2}}{4}}{4\pi \varepsilon _{0}L_{2}^{2}}}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(5)

We know that F 2 = m g tan ⁡ θ 2 {\displaystyle \mathbf {F} _{2}=mg\tan \theta _{2}}

What happens to the force between two charges when you increase the charge on one of the two charges?
and:

q 2 4 4 π ε 0 L 2 2 = m g tan ⁡ θ 2 {\displaystyle {\frac {\frac {q^{2}}{4}}{4\pi \varepsilon _{0}L_{2}^{2}}}=mg\tan \theta _{2}}

Dividing (4) by (5), we get:

( q 2 4 π ε 0 L 1 2 ) ( q 2 4 4 π ε 0 L 2 2 ) = m g tan ⁡ θ 1 m g tan ⁡ θ 2 ⇒ 4 ( L 2 L 1 ) 2 = tan ⁡ θ 1 tan ⁡ θ 2 {\displaystyle {\frac {\left({\cfrac {q^{2}}{4\pi \varepsilon _{0}L_{1}^{2}}}\right)}{\left({\cfrac {\frac {q^{2}}{4}}{4\pi \varepsilon _{0}L_{2}^{2}}}\right)}}={\frac {mg\tan \theta _{1}}{mg\tan \theta _{2}}}\Rightarrow 4{\left({\frac {L_{2}}{L_{1}}}\right)}^{2}={\frac {\tan \theta _{1}}{\tan \theta _{2}}}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(6)

Measuring the angles θ 1 {\displaystyle \theta _{1}}

What happens to the force between two charges when you increase the charge on one of the two charges?
and θ 2 {\displaystyle \theta _{2}}
What happens to the force between two charges when you increase the charge on one of the two charges?
and the distance between the charges L 1 {\displaystyle \mathbf {L} _{1}} and L 2 {\displaystyle \mathbf {L} _{2}}
What happens to the force between two charges when you increase the charge on one of the two charges?
is sufficient to verify that the equality is true taking into account the experimental error. In practice, angles can be difficult to measure, so if the length of the ropes is sufficiently great, the angles will be small enough to make the following approximation:

tan ⁡ θ ≈ sin ⁡ θ = L 2 ℓ = L 2 ℓ ⇒ tan ⁡ θ 1 tan ⁡ θ 2 ≈ L 1 2 ℓ L 2 2 ℓ {\displaystyle \tan \theta \approx \sin \theta ={\frac {\frac {L}{2}}{\ell }}={\frac {L}{2\ell }}\Rightarrow {\frac {\tan \theta _{1}}{\tan \theta _{2}}}\approx {\frac {\frac {L_{1}}{2\ell }}{\frac {L_{2}}{2\ell }}}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(7)

Using this approximation, the relationship (6) becomes the much simpler expression:

L 1 2 ℓ L 2 2 ℓ ≈ 4 ( L 2 L 1 ) 2 ⇒ L 1 L 2 ≈ 4 ( L 2 L 1 ) 2 ⇒ L 1 L 2 ≈ 4 3 {\displaystyle {\frac {\frac {L_{1}}{2\ell }}{\frac {L_{2}}{2\ell }}}\approx 4{\left({\frac {L_{2}}{L_{1}}}\right)}^{2}\Rightarrow {\frac {L_{1}}{L_{2}}}\approx 4{\left({\frac {L_{2}}{L_{1}}}\right)}^{2}\Rightarrow {\frac {L_{1}}{L_{2}}}\approx {\sqrt[{3}]{4}}}

What happens to the force between two charges when you increase the charge on one of the two charges?

(8)

In this way, the verification is limited to measuring the distance between the charges and check that the division approximates the theoretical value.

See also

  • What happens to the force between two charges when you increase the charge on one of the two charges?
    Electronics portal

  • Biot–Savart law
  • Darwin Lagrangian
  • Electromagnetic force
  • Gauss's law
  • Method of image charges
  • Molecular modelling
  • Newton's law of universal gravitation, which uses a similar structure, but for mass instead of charge
  • Static forces and virtual-particle exchange

References

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  4. ^ a b Coulomb (1785) "Premier mémoire sur l'électricité et le magnétisme," Histoire de l'Académie Royale des Sciences, pp. 569–577 — Coulomb studied the repulsive force between bodies having electrical charges of the same sign:

    Il résulte donc de ces trois essais, que l'action répulsive que les deux balles électrifées de la même nature d'électricité exercent l'une sur l'autre, suit la raison inverse du carré des distances. Translation: It follows therefore from these three tests, that the repulsive force that the two balls — [that were] electrified with the same kind of electricity — exert on each other, follows the inverse proportion of the square of the distance.

    — Coulomb (1785b) "Second mémoire sur l'électricité et le magnétisme," Histoire de l'Académie Royale des Sciences, pages 578–611

    Coulomb also showed that oppositely charged bodies obey an inverse-square law of attraction.

  5. ^ a b Purcell, Edward M. (21 January 2013). Electricity and magnetism (Third ed.). Cambridge. ISBN 9781107014022.
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    On pages 111 and 112 the author states: "We may therefore conclude that the electric attraction and repulsion must be inversely as some power of the distance between that of the 2 + 150 th and that of the 2 − 150 th, and there is no reason to think that it differs at all from the inverse duplicate ratio".
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  • Coulomb, Charles Augustin (1788) [1785]. "Premier mémoire sur l'électricité et le magnétisme". Histoire de l'Académie Royale des Sciences. Imprimerie Royale. pp. 569–577.
  • Coulomb, Charles Augustin (1788) [1785]. "Second mémoire sur l'électricité et le magnétisme". Histoire de l'Académie Royale des Sciences. Imprimerie Royale. pp. 578–611.
  • Coulomb, Charles Augustin (1788) [1785]. "Troisième mémoire sur l'électricité et le magnétisme". Histoire de l'Académie Royale des Sciences. Imprimerie Royale. pp. 612–638.
  • Griffiths, David J. (1999). Introduction to Electrodynamics (3rd ed.). Prentice Hall. ISBN 978-0-13-805326-0.
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What happens to the force between two charges when you increase the charge on one of the two charges?

  • Coulomb's Law on Project PHYSNET
  • Electricity and the Atom—a chapter from an online textbook
  • A maze game for teaching Coulomb's law—a game created by the Molecular Workbench software
  • Electric Charges, Polarization, Electric Force, Coulomb's Law Walter Lewin, 8.02 Electricity and Magnetism, Spring 2002: Lecture 1 (video). MIT OpenCourseWare. License: Creative Commons Attribution-Noncommercial-Share Alike.

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