Coulomb’s Law
By the end of this section, you will be able to:
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 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
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 \((F \propto 1/r^2)\) 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.
Recall that the gravitational force (Newton's law of gravitation) quantifies force as \(F_{s} = G \frac{mM}{r^2}.\).
The comparison between the two forces—gravitational and electrostatic—shows some similarities and differences. Gravitational force is proportional to the masses of interacting objects, and the electrostatic force is proportional to the magnitudes of the charges of interacting objects. Hence both forces are proportional to a property that represents the strength of interaction for a given field. In addition, both forces are inversely proportional to the square of the distances between them. It may seem that the two forces are related but that is not the case. In fact, there are huge variations in the magnitudes of the two forces as they depend on different parameters and different mechanisms. For electrons (or protons), electrostatic force is dominant and is much greater than the gravitational force. On the other hand, gravitational force is generally dominant for objects with large masses. Another major difference between the two forces is that gravitational force can only be attractive, whereas electrostatic could be attractive or repulsive (depending on the sign of charges; unlike charges attract and like charges repel).
Compare the electrostatic force between an electron and proton separated by \(0.530 \times 10^{-10}\) with the gravitational force between them. This distance is their average separation in a hydrogen atom.
Strategy
To compare the two forces, we first compute the electrostatic force using Coulomb's law, \(F = k \frac{| q_{1} q_{2}|}{r^2}\). We then calculate the gravitational force using Newton's universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude.
Solution
Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb's law yields
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 \(8.99 \times 10^{22} \thinspace \mathrm{m/s^2}\)(verification is left as an end-of-section problem).The gravitational force is given by Newton's law of gravitation as:
where \(G = 6.67 \times 10^{-11} \thinspace \mathrm{N \cdot m^2/kg^2}\). Here \(m\) and \(M\) represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields
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,
Discussion
This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature.
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.
What will happen to the electrostatic force when the distance between the charges is increased?
What type of force does the charges \(q_1\) and \(q_2\) exert when both have the same sign -either both positive or negative-?
Who formulated the law which states that the force between two charges Q1 and Q2 is proportional to their product divided by the separation distance r squared?
An electron at a certain distance from a charged particle is attracted with a certain force. If the distance is twice as the certain distance, how will you compare the forces?
An electron is placed in an electric field of 12.0 N/C to the right. What is the resulting force on the electron?
A proton and an electron are placed \(1.3\times 10^{-6} \; m \) apart. What is the force between them?
You can use scientific notation to express the answer.
\(N\)
The charge of the electron is negative
\(F={kq_1q_2 \over r^2}\)
\(F={(9.0\times 10^9 \; {Nm^2 \over C^2})(1.6\times10^{-19} \; C)(-1.6\times10^{-19} \; C) \over (1.3\times10^{-6} \; m)^2}\)
-1.36e-16 N
A typical AAA battery can move 2000 C of charge at 1.5 V. How long will this run a 50 mW LED?
What is the internal energy of a system consisting of two point charges, one 2.0 µC, and the other −3.0 µC, placed 1.2 m away from each other?
Which graph illustrates the relationship between electrostatic force and the distance between the charges.
Formulate a formula that states "the law of force between charges Q1 and Q2 are proportional to their product divided by the separation distance r squared."
A helium nucleus has a \(2e\) charge and a neon nucleus has a charge of \(10e\), where \(e\) is the fundamental charge, \(1.6\times10^{-19} \;C\). Find the repulsive force exerted on each of them due to the other when they are separated by 3 nm.
Convert the units of r
\(F=k {q1q2\over r^2}\)
Convert the units of r:
\(3\; nm \left( 10^{-9} \; m \over 1\;nm\right)=3\times10^{-9}\;m\)
\(F=9\times 10^9 \; {Nm^2 \over C^2} {20(1.6\times10^{-19}\;C)^2 \over (3\times10^{-9})^2}\)
Answer: 5.1e-10 N
A proton and an electron are m meters apart. What will happen to the force between them if they are more than m meters apart?
Calculate the electric force between two charges, one of \(6 \space \mu C\) and one of \(-8 \space \mu C\) separated \(1.5 \space m\).
\(F_e={(9\times10^9 \space {Nm^2 \over r^2})(6\times10^{-6} \space C)(-8\times10^{-6} \space C) \over (1.5 \space m)^2}\)
Answer: \(F_e=-0.192 \space N\)
A proton and an electron are m meters apart. What will happen to the force between them if they are less than m meters apart?
Two small spheres with the same charge, \(q=6 \; C\), experience a force of 29 N. Determine the distance between them.
\(m\)
Since the charges are the same: \(F=k{q^2\over r^2}\)
\(r=\sqrt{kq^2\over F}\)
\(r=\sqrt{(9\times 10^9 \; {Nm^2 \over C^2})(6 \; C)^2\over 29 \; N}\)
Answer: \(r=105699.6395 \; m\)
Two small spheres with the same charge, \(q=8 \times 10^{-5} \; C\), are separated by 1.9 m. Determine the force between them.
Since the charges are the same: \(F=k{q^2 \over r^2}\)
\(F={(9\times 10^9 \; {Nm^2 \over C^2})(8 \times 10^{-5} \; C)^2 \over 1.9^2}\)
Answer: \(F=15.9557 \; N\)
A proton and an electron are m meters apart. What do you think will happen to the distance between them if the force between them increases?
How much force is between the spheres based on the illustration?
\(F={(9\times 10^9 \; {Nm^2 \over C^2})(2.8\times 10^{-5} \; C)(-2.5\times 10^{-5} \; C)\over (2\times 10^{-2} \; m)^2}\)
Answer: \(F=-15750 \; N\)
A proton and an electron are m meters apart. What do you think will happen to the distance between them if the force between them decreases?
The force between a proton and an electron is \(-6 \times 10^{-21} \; N\). How far apart are the charges?
You can express the answer in scientific notation.
\(r= \sqrt{ kq_1q_2 \over F}\)
\(r= \sqrt{ (9.0\times 10^9 {Nm^2 \over C^2})(1.6\times10^{-19} \; C)(-1.6\times10^{-19} \; C) \over -6 \times 10^{-21} \; N}\)
Answer: 1.96E-4 m
Two charged objects are placed 5 m apart. Object A has a charge of \(5\times10^{-19} \; C\). How much charge does object B posses if the force between them is \(6 \times10^{-28} \; N\)?
\(C\)
\(q_B={Fr^2 \over kq_A }\)
\(q_B={(6\times 10^{-28} \; N)(5 \; m)^2 \over (9.0\times 10^9 \; {Nm^2 \over C^2})(5\times10^{-19} \; C) }\)
Answer: 3.33E-18 C
Two small spheres with the same charge are separated by 1.5 m. Determine the charge of the spheres if each experiences a 4-N force.
\(q=\sqrt{Fr^2 \over k}\)
\(q=\sqrt{(4 \; N)(1.5 \; m)^2 \over 9\times 10^9 \; {Nm^2 \over C^2}}\)
Answer: 3.16E-5 C
A helium nucleus has a charge \(q_1=+4e\) and a neon nucleus has a charge \(q_2=+5e\), where \(e=1.6 \times 10^{-19 } \; C\). Find the repulsive force experienced by the particles, considering that they are separated by \(1.9 \times10^{-9} \; m\).
You can express your answer in scientific notation.
\(F={(9\times 10^9 \; {Nm^2 \over C^2})(4e)(5e) \over (1.9\times 10^{-9} \; m)^2}\)
\(F={(9\times 10^9 \; {Nm^2 \over C^2})(20e) \over (1.9\times 10^{-9} \; m)^2}\)
\(F={20(9\times 10^9 \; {Nm^2 \over C^2})(1.6 \times 10^{-19 } \; C) \over (1.9\times 10^{-9} \; m)^2}\)
Answer: 7.98E9 N
Calculate the repulsive force between two charges, \(q_1=0.9 \; C\) and \(q_2=0.8 \; C\) , if they are separated by a distance of 1 m.
\(F={(9\times 10^9 \; {Nm^2 \over C^2})(0.9 \; C)(0.8 \; C)\over(1 \; m)^2}\)
Answer: 6.48E9 N
Two identical coins rest on a table with a separation of 2.2 m and contain identical charges. How large is the charge in each coin if one experiences a force of 1 N?
The diameter of the coins is small compared to the separation between them, therefore we can assume that the charges are punctual.
\(F=k{q_1q_2\over r^2}\)
Since the coins are identical, \(q_1=q_2=q \; \Rightarrow \; F=k{q^2\over r^2}\)
\(q=\sqrt{Fr^2\over k}\)
\(q=\sqrt{(1 \;N)(2.2 \; m)^2\over 9\times 10^9 \; {Nm^2 \over C^2}}\)
Answer: 2.32E-5 C
Two identical coins rest submerged in a tub with water with a separation of 2.3 m and contain identical charges. How large is the charge in each coin if one experiences a force of 8 N?
The diameter of the coins is small compared to the separation between them, therefore we can assume that the charges are punctual. Consider the dielectric constant of water \(K=80\).
\(F={k\over K}{q_1q_2\over r^2}\)
\(q=\sqrt{KFr^2\over k}\)
\(q=\sqrt{80(8 \;N)(2.3 \; m)^2)\over 9\times 10^9 \; {Nm^2 \over C^2}}\)
Answer: 6.13E-4 C
A glass bar charged with \(Q=\)2 C is brought close to a piece of paper at a distance of 7 mm exerting an electrical force \(F=-4.27 \times 10^{16}\;N\). Compute for the charge q induced in the piece of paper.
\(F=K{Q \cdot q \over d^2} \rightarrow q={F \cdot d^2 \over k \cdot Q}\)
Convert the units of \(d\): \(d=7\;mm({1\;m \over 1000\;mm})=0.007\;m\)
Answer: \(q=116.2079\;C\)
A charged particle \(P_1\) has moved 39 cm away from another charged particle \(P_2\), both have the same sign. If a work W=34 J has been done to move \(P_1\), what is the amount of the repulsion force between the particles?
\(F={W \over r}\)
Convert the units of \(r\): \(r=39\;cm({1\;m \over 100\;cm})=0.39\;m\)
Answer: \(F=87.1795\;N\)
Calculate the work required to separate two charged particles by a distance of r=5 cm, if the electric force between them is F=31 N.
\(J\)
\(F={W \over r} \rightarrow W=F \cdot r\)
Convert the units of \(r\): \(r=5\;cm({1\;m \over 100\;cm})=0.05\;m\)
Answer: \(W=1.55\;J\)
Calculate the work needed to bring together two charges \(q_1 = 6\;C\) and \(q_2 = 6\;C\) at distance \(r = 1.77\;m\).
\(W={q_1 \cdot q_2 \over r}\)
Answer: \(W=20.339\;J\)
Solve for the electric force.
Answer: \(F=11.491\;N\)
Two charged particles \(q_1 = 2.5 \;\mu C\) and \(q_2 = -8.6 \;\mu C\) are \(r = 13.46\;cm\) apart.
a. Calculate the force between the particles in SI units.
\(F = K {q_1 \cdot q_2 \over r^2}\)
Convert the units of r: \(r = 13.46\;\not cm \bigg( {1\;m \over 100\;\not cm} \bigg) = 1.35 \times 10^{-1}\;m\)
Convert the units of q:
\(\;\;\;\;\;\;\;\;\;\;q_1 = 2.5\;\not \mu C \bigg( {1\;C \over 1 \times 10^6\;\not \mu C} \bigg) = 2.5 \times 10^{-6}\;C \)
\(\;\;\;\;\;\;\;\;\;\;q_2 = -8.6\;\not \mu C \bigg( {1\;C \over 1 \times 10^6\;\not \mu C} \bigg) = -8.6 \times 10^{-6}\;C \)
In SI units, \(K = 9.0 \times 10^9 {N \cdot m^2 \over C^2}\);
\(F = 9.0 \times 10^9 {N \cdot m^2 \over C^2} {(2.5 \times 10^{-6}\;C)(-8.6 \times 10^{-6}\;C) \over (1.35 \times 10^{-1}\;m)^2}\)
Answer: \(F = -10.6805\;N\)
b. Calculate the force between the particles in cgs units.
\(dyn\)
\(F = {q_1 \cdot q_2 \over r^2}\)
\(\;\;\;\;\;\;\;\;\;\;q_1 = 2.5\;\not \mu C \bigg( {1\;C \over 1 \times 10^6\;\not \mu C} \bigg) \bigg( {3.0 \times 10^9\;esu \over 1\;\not C} \bigg)= 7.5 \times 10^{3}\;esu\)
\(\;\;\;\;\;\;\;\;\;\;q_2 = -8.6\;\not \mu C \bigg( {1\;C \over 1 \times 10^6\;\not \mu C} \bigg) \bigg( {3.0 \times 10^9\;esu \over 1\;\not C} \bigg)= -2.58 \times 10^{4}\;esu\)
Answer: \(F=\)-1.07E6 dyn
What is the electrical force between an \(\alpha - particle\) and a proton if they are 3 Å away?
\(F = {q_{\alpha} \cdot q_p \over r^2}\)
Convert the units of r: \(r = 3\;\not {A^\circ} \bigg( {1\;m \over 10^{10}\;\not {A^\circ}} \bigg) = 3 \times 10^{-10}\;m\)
Answer: \(F =\) 5.69E-19 N