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## High school physics - NGSS

Course: high school physics - ngss > unit 2.

- Coulomb's law
- Coulomb's law and electric force review

## Understand: Coulomb's law

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## 2.1: Coulomb’s Law

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learning objectives

- Apply the superposition principle to determine the net response caused by two or more stimuli

The superposition principle (also known as superposition property) states that: for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually . For Coulomb’s law, the stimuli are forces. Therefore, the principle suggests that total force is a vector sum of individual forces.

## Coulomb Force

The scalar form of Coulomb’s Law relates the magnitude and sign of the electrostatic force F, acting simultaneously on two point charges q 1 and q 2 :

\[| \mathrm { F } | = \dfrac { 1 } { 4 \pi a r \epsilon _ { 0 } } \dfrac { \left| q _ { 1 } q _ { 2 } \right| } { \mathrm { r } ^ { 2 } }\]

Lorentz Force on a Moving Particle : Lorentz force f on a charged particle (of charge q) in motion (instantaneous velocity v). The E field and B field vary in space and time.

where r is the separation distance and ε 0 is electric permittivity. If the product q 1 q 2 is positive, the force between them is repulsive; if q 1 q 2 is negative, the force between them is attractive. The principle of linear superposition allows the extension of Coulomb’s law to include any number of point charges—in order to derive the force on any one point charge by a vector addition of these individual forces acting alone on that point charge. The resulting force vector happens to be parallel to the electric field vector at that point, with that point charge removed.

To calculate the force on a small test charge q at position rr, due to a system of N discrete charges:

\[\mathrm{F(r)=\dfrac{q}{4πarϵ_0} }\sum _ { i = 1 } ^ { N } q _ { i } \frac { \mathrm { r } - \mathrm { r } _ { \mathrm { i } } } { \left| \mathrm { r } - \mathrm { r } _ { \mathrm { i } } \right| ^ { 3 } } = \frac { q } { r 4 \pi a r \epsilon _ { 0 } }\sum _ { \mathrm { i } = 1 } ^ { \mathrm { N } } \mathrm { q } _ { \mathrm { i } } \dfrac { \hat { \mathrm { R } _ { i } } } { \left| \mathrm { R } _ { \mathrm { i } } \right| ^ { 2 } }\]

where q i and r i are the magnitude and position vector of the i-th charge, respectively, and \(\hat { \mathrm { R } _ { \mathrm { i } } }\) is a unit vector in the direction of \(\mathrm { R } _ { \mathrm { i } } = \mathrm { r } - \mathrm { r } _ { \mathrm { i } }\) (a vector pointing from charges q i to q. )

Of course, our discussion of superposition of forces applies to any types (or combinations) of forces. For example, when a charge is moving in the presence of a magnetic field as well as an electric field, the charge will feel both electrostatic and magnetic forces. Total force, affecting the motion of the charge, will be the vector sum of the two forces. (In this particular example of the moving charge, the force due to the presence of electromagnetic field is collectively called Lorentz force (see ).

## Spherical Distribution of Charge

The charge distribution around a molecule is spherical in nature, and creates a sort of electrostatic “cloud” around the molecule.

- Describe shape of a Coulomb force from a spherical distribution of charge

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.

Charge distribution in a water molecule : Schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving a permanent charge separation as shown. Water is thus a polar molecule. It is more easily affected by electrostatic forces than molecules with uniform charge distributions.

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 ) to an accuracy of 1 part in 1016. No exceptions have ever been found, even at the small distances within the atom.

Coulomb’s law holds even within the atoms, correctly describing the force between the positively charged 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 energy of attraction approaches zero and ionic bonding is less favorable. As the magnitude of opposing charges increases, energy increases and ionic bonding is more favorable.

An electric field is a vector field which associates to each point of the space the Coulomb force that will experience a test unity charge. Given the electric field, the strength and direction of a force F on a quantity charge q in an electric field E is determined by the electric field. For a positive charge, the direction of the electric field points along lines directed radially away from the location of the point charge, while the direction is towards for a negative charge.

This distribution around a charged molecule is spherical in nature, and creates a sort of electrostatic “cloud” around the molecule. The attraction or repulsion forces within the spherical distribution of charge is stronger closer to the molecule, and becomes weaker as the distance from the molecule increases.

This image shows the outer electron cloud of a neutral water molecule. The charge distribution of the oxygen molecule is negative, and attracts the two positive hydrogen molecules. The attraction between the two opposing charges forms a neutral water molecule. It is a polar molecule because there is still a permanent charge separation because the electrons spend more time near the oxygen than the hydrogens.

## Solving Problems with Vectors and Coulomb’s Law

Coulomb’s Law, which calculates the electric force between charged particles, can be written in vector notation as \(\mathrm { F } ( \mathrm { E } ) = \frac { \mathrm { kq } _ { 1 } \mathrm { q } _ { 2 } } { \mathrm { r } ^ { 2 } } \mathrm { r } +\).

- Explain when the vector notation of Coulomb’s Law can be used

## Electric Force Between Two Point Charges

To address the electrostatic forces among electrically charged particles, first consider two particles with electric charges q and Q , separated in empty space by a distance r. Suppose that we want to find the electric force vector on charge q. (The electric force vector has both a magnitude and a direction. ) We can express the location of charge q as r q , and the location of charge Q as r Q . In this way we can know both how strong the electric force is on a charge, but also what direction that force is directed in. Coulomb’s Law using vectors can be written as:

\[\mathbf { F } _ { \mathbf { E } } = \dfrac { \operatorname { kq } Q \left( \mathrm { r } _ { \mathrm { q } } - \mathrm { r } _ { Q } \right) } { \left| \mathrm { r } _ { \mathrm { q } } - \mathrm { r } _ { Q } \right| ^ { 3 } }\]

In this equation, k is equal to \(\frac { 1 } { 4 \pi \varepsilon _ { 0 } \varepsilon }\) ,where \(\varepsilon _ { 0 }\) is the permittivity of free space and εε is the relative permittivity of the material in which the charges are immersed. The variables \(\mathbf { F} _ { \mathbf { E } } , \mathbf { \Gamma } _ { \mathrm { q } }\) and \(\mathbf{ R}_Q\) are in bold because they are vectors. Thus, we need to find \(\mathbf { r } _ { \mathrm { q } } - \mathbf { r } _ { \mathrm { Q } }\) by performing standard vector subtraction. This means that we need to subtract the corresponding components of vector\(\mathbf{r}_\mathrm{Q}}\) from vector \(\mathrmbf{r}_\mathrm{q}\).

This vector notation can be used in the simple example of two point charges where only one of which is a source of charge.

Application of Coulomb’s Law : In a simple example, the vector notation of Coulomb’s Law can be used when there are two point charges and only one of which is a source charge.

## Electric Force on a Field Charge Due to Fixed Source Charges

Suppose there is more than one point source charges providing forces on a field charge. diagrams a fairly simple example with three source charges (shown in green and indexed by subscripts) and one field charge (in red, designated q). We assume that the source charges are fixed in space, and the field charge q is subject to forces from the source charges.

Multiple point charges : Coulomb’s Law applied to more than one point source charges providing forces on a field charge.

Note the coordinate system that has been chosen. All of the charges lie on the corners of a square, and the origin is chosen to collocate with the lower right source charge, and aligned with the square. Since we can have only one origin of coordinates, no more than one of the source points can lie at the origin, and the displacements from different source points to the field point differ. The total force on the field charge q is due to applications of the force described in the vector notation of Coulomb’s Law from each of the source charges. The total force is therefore the sum of these individual forces.

Displacements of field charge : The displacements of the field charge from each source charge are shown as light blue arrows.

Applying Coulomb’s Law three times and summing the results gives us:

\[\mathbf { F } _ { \mathbf { E } _ { \mathbf { q } } } =\dfrac { \operatorname { kq } \cdot q _ { 1 } \left( \mathbf { r } _ { q } - \mathbf { r } _ { q 1 } \right) } { \left| \mathbf { r } _ { \mathrm { q } } - \mathbf { r } _ { \mathrm { q } 1 } \right| ^ { 3 } }+ \dfrac { \mathrm { kq } \cdot \mathbf { q } _ { 2 } \left( \mathbf { r } _ { \mathrm { q } } - \mathbf { r } _ { \mathrm { q } 2 } \right) } { \left| \mathbf { r } _ { \mathrm { q } } - \mathbf { r } _ { \mathrm { q } 2 } \right| ^ { 3 } }+ \dfrac { \mathrm { kq } \cdot \mathrm { q } _ { 3 } \left( \mathbf { r } _ { \mathrm { q } } - \mathbf { r } _ { \mathrm { q } 3 } \right) } { \left| \mathbf { r } _ { \mathrm { q } } - \mathbf { r } _ { \mathrm { q } 3 } \right| ^ { 3 } }\]

This equation can further be simplified and applied to a fixed number of charge points.

\[\mathbf { F } _ { \mathbf { n } } =\sum _ { i \neq n }\dfrac { q _ { n } q _ { i } \left( \mathbf { r } _ { n } - \mathbf { r } _ { i } \right) } { 4 \pi \epsilon _ { 0 } \left| \mathbf { r } _ { \mathrm { n } } - \mathbf { r } _ { \mathrm { i } } \right| ^ { 3 } }\]

Coulomb’s Law : In this video I continue with my series of tutorial videos on Electrostatics. It’s pitched at undergraduate level and while it is mainly aimed at physics majors, it should be useful to anybody taking a first course in electricity and magnetism such as engineers etc.

- The superposition principle suggests that the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually.
- Total Coulomb force on a test charge due to a group of charges is equal to the vector sum of all the Coulomb forces between the test charge and other individual charges.
- The superposition of forces is not limited to Coulomb forces. It applies to any types (or combinations) of forces.
- The force between two objects is inversely proportional to the square of the distance between two objects.
- The attraction or repulsion forces within the spherical distribution of charge is stronger closer to the molecule and becomes weaker as the distance from the molecule increases.
- This law also 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.
- The vector notation of Coulomb ‘s Law can be used in the simple example of two point charges where only one of which is a source of charge.
- The total force on the field charge for multiple point source charges is the sum of these individual forces.
- Coulomb’s Law can be further simplified and applied to a fixed number of charge points.
- Lorentz force : The force exerted on a charged particle in an electromagnetic field.
- unit vector : A vector with length 1.
- electrostatic force : The electrostatic interaction between electrically charged particles; the amount and direction of attraction or repulsion between two charged bodies.
- coulomb’s law : the mathematical equation calculating the electrostatic force vector between two charged particles

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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

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

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 Figure 18.19 .)

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 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 10 16 size 12{"10" rSup { size 8{"16"} } } {} . No exceptions have ever been found, even at the small distances within the atom.

## Example 18.1

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 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.

To compare the two forces, we first compute the electrostatic force using Coulomb’s law, F = k | q 1 q 2 | r 2 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.

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

Thus the Coulomb force is

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 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:

where G = 6.67 × 10 − 11 N ⋅ m 2 / kg 2 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 m and M M represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields

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,

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.

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.

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- Authors: Paul Peter Urone, Roger Hinrichs
- Publisher/website: OpenStax
- Book title: College Physics
- Publication date: Jun 21, 2012
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- Section URL: https://openstax.org/books/college-physics/pages/18-3-coulombs-law

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## Module 6 – Electricity

Module 6 coulomb’s law, learning objectives.

By the end of this section, you will be able to:

- 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.

Figure 1. 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

[latex]\displaystyle{F}=k\frac{\mid{q}_1q_2\mid}{r^2}\\[/latex]

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

[latex]\displaystyle{k}=8.988\times10^9\frac{\text{N}\cdot\text{m}^2}{\text{C}^2}\approx8.99\times10^9\frac{\text{N}\cdot\text{m}^2}{\text{C}^2}\\[/latex]

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 Figure 2).

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 [latex]\left(F\propto\frac{1}{r^2}\right)\\[/latex] to an accuracy of 1 part in 10 16 . No exceptions have ever been found, even at the small distances within the atom.

Figure 2. The magnitude of the electrostatic force F between point charges q 1 and q 2 separated by a distance r 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 is equal in magnitude and opposite in direction to the force it exerts on q 2 . (a) Like charges. (b) Unlike charges.

## Example 1. 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 with the gravitational force between them. This distance is their average separation in a hydrogen atom.

To compare the two forces, we first compute the electrostatic force using Coulomb’s law, [latex]\displaystyle{F}=k\frac{\mid{q}_1q_2\mid}{r^2}\\[/latex]. 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.

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

[latex]\begin{array}{lll}{F}&=&k\frac{\mid{q}_1q_2\mid}{r^2}\\\text{ }&=&8.99\times10^9\frac{\text{N}\cdot\text{m}^2}{\text{C}^2}\times\frac{\left(1.60\times10^{-19}\text{ C}\right)\left(1.60\times10^{-19}\text{ C}\right)}{\left(0.530\times10^{-10}\text{ m}\right)^2}\end{array}\\[/latex]

Thus the Coulomb force is F = 8.19 × 10 −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 (verification is left as an end-of-section problem).The gravitational force is given by Newton’s law of gravitation as:

[latex]F_G=G\frac{mM}{r^2}\\[/latex],

where G = 6.67 × 10 −11 N · m 2 /kg 2 . Here m and M represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields

[latex]F_G=\left(6.67\times10^{-11}\text{N}\cdot\text{m}^2\text{/kg}^2\right)\times\frac{\left(9.11\times10^{-31}\text{ kg}\right)\left(1.67\times10^{-27}\text{ kg}\right)}{\left(0.530\times10^{-10}\text{ m}\right)^2}=3.61\times10^{-47}\text{ N}\\[/latex]

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, [latex]\frac{F}{F_G}=2.27\times10^{39}\\[/latex].

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.

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 [latex]F=k\frac{\mid{q}_{1}{q}_{2}\mid}{{r}^{2}}\\[/latex], where q 1 and q 2 are two point charges separated by a distance r , and [latex]k\approx8.99\times{10}^{9}\frac{\text{N}\cdot {\text{m}}^{2}}{{\text{C}}^{2}}\\[/latex]
- 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

Figure 3. Schematic representation of the outer electron cloud of a neutral water molecule.

Use Figure 3 as a reference in the following questions. Figure 3 shows a schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving a permanent charge separation as shown. Water is thus a polar molecule . It is more easily affected by electrostatic forces than molecules with uniform charge distributions.

- Figure 3 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 Figure 3, explain, in terms of Coulomb’s law, why a polar molecule (such as in Figure 3) 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 μ C charge and a silk cloth with a –0.600 μ 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.
- 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?
- How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?
- If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge.
- A test charge of +2 μ C is placed halfway between a charge of +6 μ C and another of +4 μ C separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6 μ C charge)?
- Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
- (a) By what factor must you change the distance between two point charges to change the force between them by a factor of 10? (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force.
- Suppose you have a total charge q tot that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force?
- (a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece’s weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 10.0 mg piece of tape held 1.00 cm above another. (b) Discuss whether the magnitude of this charge is consistent with what is typical of static electricity.
- (a) Find the ratio of the electrostatic to gravitational force between two electrons. (b) What is this ratio for two protons? (c) Why is the ratio different for electrons and protons?
- At what distance is the electrostatic force between two protons equal to the weight of one proton?
- A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms’ electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons.
- (a) Two point charges totaling 8.00 µ C exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the charge on each? (b) What is the charge on each if the force is attractive?
- Point charges of 5.00 µ C and –3.00 µ C are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?
- Two point charges q 1 and q 2 are 3.00 m apart, and their total charge is 20 µ C. (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer.

Coulomb’s law: the mathematical equation calculating the electrostatic force vector between two charged particles

Coulomb force: another term for the electrostatic force

electrostatic force: the amount and direction of attraction or repulsion between two charged bodies

## Selected Solutions to Problems & Exercises

2. (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.

4. The separation decreased by a factor of 5.

8. [latex]\begin{array}{lll}F&=&k\frac{\mid{q}_{1}{q}_{2}\mid}{{r}^{2}}=ma\Rightarrow{a}=\frac{k{q}^{2}}{mr^{2}}\\\text{ }&=&\frac{\left(9.00\times{10}^{9}\text{ N}\cdot\text{m}^{2}/{\text{C}}^{2}\right){\left(1.60\times {10}^{-19}\text{ m}\right)}^{2}}{\left(1.67\times {10}^{-27}\text{ kg}\right){\left(2.00\times {10}^{-9}\text{ m}\right)}^{2}}\\ & =& 3.45\times {10}^{16}\text{ m/s}^{2}\end{array}\\[/latex]

9. (a) 3.2; (b) If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10.

11. (a) 1.04 × 10 −9 C; (b) This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity

14. 1.02×10 −11

16. (a) 0.859 m beyond negative charge on line connecting two charges; (b) 0.109 m from lesser charge on line connecting two charges

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## Coulomb's Law

- Static Electricity Tutorial - Coulomb's Law
- Static Electricity Tutorial - Inverse Square Law
- The Curriculum Corner : Coulomb's Law

## Search the NGSS Corner

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## Static Electricity

Electric fields, 7th - 10th .

## Coulomb's Law

20 questions

According to Coulomb's Law, when the distance increases, the electrostatic force ____________.

remains unchanged

has no relationship to distance

The Coulomb is the unit for:

What happens to the force between two charged objects when you double the distance between them?

it is cut in half

it quadruples

it is cut by 1/4

If one of the charges doubles, what happens to the force?

it gets cut in half

What is the unit for electrostatic charge?

Coulomb’s law states that the force between two charged objects will __________ when the magnitude (value) of the object’s charge increases.

stay the same

first increase, then decrease

What would a negative value for F e mean?

The particles are moving away from each other

The particles have the same charge

The particles have opposite charges

The particles are moving towards each other

Select all answer that apply:

What kinds of particles do Coulomb's Law apply to?

neutral atoms

A repulsive force exists between which two particles?

Two neutrons

An electron and a neutron

An electron and a proton

Two protons

The variable "r" in Coulomb's Law would have what units?

Which is the value of the constant in Coulomb's equation?

k e = 9·10 -9

k e = 6.67·10 -11

k e = 9 ·10 9

What does "r" measure?

the distance between the two charged objects

the amount of charge carried by the objects

the amount of force between two charged objects

the constant value for forces between the objects

What does F e stand for?

the electrostatic force

the symbol for iron

the constant for the charges

the charges of the particles

What will be the units of F (or F e )?

As the value of r increases, what happens to the value of F?

F increases

F decreases

F never changes, q does

Cannot know from the information

What does q measure?

the amount of charge carried by the object

Why are there two q's in the equation?

because there are two charged objects

because there are protons and electrons

because the particles are moving away from each other

because the particles are moving towards each other

What can you do to increase the value of k?

change the force

change the distance

change the charge

nothing can be done to change the value of k; it is a constant

An attractive force exists between which two particles?

Two electrons

Coulomb's Law is __________ proportional.

polymetrically

exponentially

- IIT JEE Study Material
- Coulombs Law

## Coulomb's Law

The force between charged bodies is no contact force. It exists over a length, and all electrical interaction has a force embedded in it. The charges and distance between the charged bodies are the factors that determine the power and influence of the force. The same force exists, whether it’s a plastic comb attracting paper pieces or two electrons repelling each other.

Download Complete Chapter Notes of Electric Charges and Fields Download Now

## What Is Coulomb’s Law?

According to Coulomb’s law , the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges.

## Table of Content

Coulomb’s law formula, coulomb’s law in vector form.

- What Is One Coulomb of Charge
- Limitations
- Relative Permittivity
- Applications

In short, F ∝ q 1 q 2 /d 2

- ε is absolute permittivity ,
- K or ε r is the relative permittivity or specific inductive capacity
- ε 0 is the permittivity of free space .
- K or ε r is also called a dielectric constant of the medium in which the two charges are placed.

## History of Coulomb’s Law

In 1785, French physicist Charles Augustin de Coulomb coined a tangible relationship in mathematical form between two bodies that have been electrically charged. He published an equation for the force causing the bodies to attract or repel each other, which is known as Coulomb’s law or Coulomb’s inverse-square law .

Here, F 12 is the force exerted by q 1 on q 2, and F 21 is the force exerted by q 2 on q 1 .

Coulomb’s law holds for stationary charges only, which are point sized. This law obeys Newton’s third law

Force on a charged particle due to a number of point charges is the resultant of forces due to individual point charges, i.e.,

## What Is 1 Coulomb of Charge?

A Coulomb is a charge which repels an equal charge of the same sign with a force of 9×10 9 N when the charges are one metre apart in a vacuum. Coulomb force is the conservative mutual and internal force.

The value of ε o is 8.86 × 10 -12 C 2 /Nm 2 (or) 8.86 × 10 -12 Fm –1

Note: Coulomb force is true only for static charges .

## Coulomb’s Law – Conditions for Stability

If q is slightly displaced towards A, F A increases in magnitude while F B decreases in magnitude. Now, the net force on q is toward A, so it will not return to its original position. So for axial displacement, the equilibrium is unstable.

If q is displaced perpendicular to AB, the force F A and F B bring the charge to its original position. So, for perpendicular displacement, the equilibrium is stable.

## Key Points on Coulomb’s Law

2. Kr 2 = constant or K 1 r 1 2 = K 2 r 2 2

3. If the force between two charges separated by a distance ‘r 0 ’ in a vacuum is the same as the force between the same charges separated by a distance ‘r’ in a medium, then from Coulomb’s law, Kr 2 = r 0 2

4. Two identical conductors having charges q 1 and q 2 are put to contact and then separated, after which each will have a charge equal to (q 1 + q 2 )/2. If the charges are q 1 and –q 2 , each will have a charge equal to (q 1 – q 2 )/2.

5. Two spherical conductors having charges q 1 and q 2 and radii r 1 and r 2 are put to contact, and then separated the charges of the conductors after contact is

q 1 = [r 1 /(r 1 + r 2 )] (q 1 + q 2 ) and q 2 = [r 2 /(r 1 + r 2 )] (q 1 + q 2 )

7. If charges are q 1 and -q 2, then F = F(q 1 + q 2 ) 2 / 4q 1 q 2

8. Between two electrons separated by a certain distance: Electrical force/Gravitational force = 10 42

9. Between two protons separated by a certain distance: Electrical force/Gravitational force = 10 36

10. Between a proton and an electron separated by a certain distance: Electrical force/Gravitational force = 10 39

11. The relationship between the velocity of light, the permeability of free space and the permittivity of free space is given by the expression c = 1 / √ (μ o ε o )

12. If Coulomb’s law is applied to two identical balls of mass m are hung by silk thread of length ‘l’ from the same hook and carry similar charges q, then,

- The distance between balls \(\begin{array}{l}=[\frac{q^{2}2l }{4\pi \epsilon _{o}mg}]^{\frac{1}{3}}\end{array} \)
- The tension in the thread \(\begin{array}{l}=\sqrt{f^2+(mg)^2}\end{array} \)
- If the total system is kept in space, then the angle between threads is 180°, and tension in a thread is given by
- A charge Q is divided into q and (Q – q). Then, the electrostatic force between them is maximum when

## Limitations of Coulomb’s Law

- The law is applicable only for the point charges at rest.
- Coulomb’s law can only be applied in those cases where the inverse square law is obeyed.
- It is difficult to implement Coulomb’s law where charges are in arbitrary shape because, in such cases, we cannot determine the distance between the charges.
- The law can’t be used directly to calculate the charge on big planets.

## Relative Permittivity of a Material

- For air K = 1
- For metals, K = infinity

The force between 2 charges depends on the nature of the intervening medium, whereas gravitational force is independent of the intervening medium.

The value of 1/4 π ε 0 is equal to 9 × 10 9 Nm 2 /C 2 .

## ⇒ Related Topics

- Electrostatics
- Electric Charge

## Application of Coulomb’s Law

- To calculate the distance and force between the two charges
- The electric field can be calculated using Coulomb’s law

Where, E = Strength of the electric field

F = Electrostatic force

Q T = Test charge in Coulombs

- To calculate the force on one point due to the presence of several points (Theorem of superposition)

## Problems on Coulomb’s Law

Problem 1: Charges of magnitude 100 microcoulomb each are located in a vacuum at the corners A, B and C of an equilateral triangle measuring 4 meters on each side. If the charge at A and C are positive and the charge at B negative, what is the magnitude and direction of the total force on the charge at C?

The situation is shown in the figure below. Let us consider the forces acting on C due to A and B.

Now, from Coulomb’s law, the force of repulsion on C due to A, i.e., FCA in the direction of AC, is given by

The force of attraction on C due to B, i.e., F CB in direction CB, is given by

Thus, the two forces are equal in magnitude. The angle between them is 120º. The resultant force F is given by

This force is parallel to AB.

Problem 2: The negative point charges of unit magnitude and a positive point charge q are placed along the straight line. At what position and for what value of q will the system be in equilibrium? Check whether it is stable, unstable or neutral equilibrium.

The two negative charges – A and B – of unit magnitude are shown in the figure below. Let the positive charge q be at a distance r A from A and at a distance r B from B.

Now, from Coulomb’s law, force on q due to A

Force on q due to B

These two forces acting on q are opposite and collinear. For the equilibrium of q, the two forces must also be equal, i.e.,

|F qA | = |F qB |

Hence, rA = rB

So, for the equilibrium of q, it must be equidistant from A & B, i.e., in the middle of AB.

Now for the equilibrium of the system, A and B must be in equilibrium.

For the equilibrium of A,

Force on A by q \(\begin{array}{l}=\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}}\end{array} \) towards q

Force on A by B \(\begin{array}{l}=\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{(1)(1)}{{{({{r}_{A}}+{{r}_{B}})}^{2}}}\end{array} \) \(\begin{array}{l}=\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{1}{{{(2{{r}_{A}})}^{2}}}\end{array} \) away from q

The two forces are opposite and collinear. For equilibrium, the forces must be equal, opposite and collinear. Hence,

or q = 1/4 in the magnitude of either charge

It can also be shown that for the equilibrium of B, the magnitude of q must be 1/4 of the magnitude of either charge.

Problem 3: A positive charge of 6×10 -6 C is 0.040m from the second positive charge of 4×10 -6 C. Calculate the force between the charges.

q 1 = 6×10 -6 C

q 2 = 4×10 -6 C

r = 0.040 m

F e = 134.85 N

Problem 4: Two-point charges, q 1 = +9 μC and q 2 = 4 μC, are separated by a distance r = 12 cm. What is the magnitude of the electric force?

k = 8.988 x 10 9 Nm 2 C −2

q 1 = 9 ×10 -6 C

q 2 = 4 ×10 -6 C

r = 12cm = 0.12 m

F e = 22.475 N

## Coulomb’s Law and Superposition Principle – Electrostatics – JEE Main 2023

## Coulomb’s Law – Concepts and Questions

## Frequently Asked Questions on Coulomb’s Law

State coulomb’s inverse-square law in electrostatics..

The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

## What are the limitations of Coulomb’s law?

Coulomb’s law holds good for stationary point charges. Coulomb’s law is not universal, as it depends on the properties of the intervening medium.

## Is the electrostatic force between two point charges a central force?

Yes. The electrostatic force between two point charges always acts along the line joining the two charges. Hence, it is a central force.

## What is one Coulomb of charge?

One Coulomb of charge is that charge which, when placed at rest in a vacuum at a distance of one metre from an equal and similar stationary charge, repels it and is repelled by it with a force of 9 x 10 9 Newton.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

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Coulomb's Law Assignment Flashcards Flashcards | Quizlet Coulomb's Law Assignment Flashcards 4.7 (51 reviews) Charges of +5C and −9C are at a distance of 1m from each other. Which diagram represents the force between the two charges? Image for option 1 Image for option 2 Image for option 3 Image for option 4 Click the card to flip 👆 image 3

Coulomb's Law Assignment Flashcards | Quizlet Coulomb's Law Assignment 5.0 (6 reviews) Charges of +5C and −9C are at a distance of 1m from each other. Which diagram represents the force between the two charges? Click the card to flip 👆 C. The arrows point towards each other and are the same length. Click the card to flip 👆 1 / 4 Flashcards Learn

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4. The electrical force is 8.0 × 10^-37 times greater than the gravitational force. 2. The electrical force is 1.2 × 10^36 times greater than the gravitational force. Study with Quizlet and memorize flashcards containing terms like Charges of +5C and −9C are at a distance of 1m from each other. Which diagram represents the force between the ...

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Coulomb's law describes the strength of the electrostatic force (attraction or repulsion) between two charged objects. The electrostatic force is equal to the charge of object 1 times the charge of object 2, divided by the distance between the objects squared, all times the Coulomb constant (k). Questions Tips & Thanks

Both forces act along the imaginary line joining the objects. Both forces are inversely proportional to the square of the distance between the objects, this is known as the inverse-square law. Also, both forces have proportionality constants. F g uses G and F E uses k , where k = 9.0 × 10 9 N ⋅ m 2 C 2 .

Understand: Coulomb's law. The body of a bee carries a charge of − 1 × 10 − 9 C after flying through the air. The charge on the outside of a granule of pollen is 1 × 10 − 11 C . The bee lands so that it is 8 × 10 − 3 m from the granule of pollen. Calculate the magnitude of the force between these two charged objects.

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 ...

Coulomb's law inverse-square law More than 100 years before Thomson and Rutherford discovered the fundamental particles that carry positive and negative electric charges, the French scientist Charles-Augustin de Coulomb mathematically described the force between charged objects.

1. 53 N 2. 3 degrees Why can scientists ignore the gravitational force when studying the physics of an atom? Sample response: Scientists can ignore the gravitational force because the electrical force between protons and electrons in the atom is millions of times stronger than the gravitational force between the protons and electrons.

The superposition principle (also known as superposition property) states that: for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually.For Coulomb's law, the stimuli are forces. Therefore, the principle suggests that total force is a vector sum of individual forces.

Figure 3.3.1 3.3. 1: The electrostatic force F F → between point charges q1 q 1 and q2 q 2 separated by a distance |r 12| = |r 21| | r → 12 | = | r → 21 | 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 q1 q 1 is equal in magnitude and ...

Figure 1: (a) Coulomb's torsion balance: A pith ball (lower right corner) is attached on a rotating beam with a counterweight on the opposite end. When a second pith ball (upper left corner) of equal charge is brought near the rst ball, they will repel, and the beam starts to rotate (Source: Coulomb, 1785). (b) A pith ball electroscope.

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. Charge distribution in a water molecule: Schematic representation of the outer electron cloud of a neutral water molecule.

Figure 5.14 The electrostatic force F → F → between point charges q 1 q 1 and q 2 q 2 separated by a distance r 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 q 1 is equal in magnitude and opposite in direction to the force it exerts ...

18.3 Coulomb's Law. Figure 18.18 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 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∣q1q2∣ r2 F = k ∣ q 1 q 2 ∣ r 2

DEIB in STEM Ed. Donate. Visualize the electrostatic force that two charges exert on each other. Observe how changing the sign and magnitude of the charges and the distance between them affects the electrostatic force.

Physics Interactives - Coulomb's Law Law of Electrostatic Force - Student Activity Grade Level: High School Description: This activity is intended for use in the early stages of a learning cycle on Electrostatics and Coulomb's Law. How is electrostatic force related to magnitude of charge and to separation distance between charged objects?

1. Multiple Choice 30 seconds 1 pt According to Coulomb's Law, when the distance increases, the electrostatic force ____________. increases decreases remains unchanged has no relationship to distance 2. Multiple Choice 30 seconds 1 pt The Coulomb is the unit for: Charge Force Distance Mass 3. Multiple Choice 30 seconds 1 pt

A Coulomb is a charge which repels an equal charge of the same sign with a force of 9×10 9 N when the charges are one metre apart in a vacuum. Coulomb force is the conservative mutual and internal force. The value of εo is 8.86 × 10-12 C2/Nm2 (or) 8.86 × 10-12 Fm-1. Note: Coulomb force is true only for static charges.