How strong is the push?
The last lesson said only that like charges repel and unlike charges attract. Now we want a number. Take two small charged balls whose size is tiny compared with the distance between them. Such a body is a point charge. Two things decide the force between two point charges: how much charge each carries, and how far apart they are.
Experiments give a simple answer. Double either charge and the force doubles. Double the distance and the force falls to a quarter. The force therefore grows as the product of the charges and falls as the square of the distance. It acts along the line joining the two charges. This is Coulomb's law.
Two point charges \(q_1\) and \(q_2\) at rest, a distance \(r\) apart in vacuum, exert on each other forces of size \(F = k|q_1q_2|/r^2\). The forces act along the line joining the charges; they repel for like charges and attract for unlike charges.
The constant is written in two ways. \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\) is the quick form. The same constant is also written \(1/4\pi\varepsilon_0\), where \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\) is the permittivity of free space. The \(4\pi\) looks odd here, but it makes later formulas for sheets and capacitors simpler. The two rounded values differ by half a percent, so each example in this book uses one of them and says which.
The law has the same shape as Newton's law of gravitation, \(F = Gm_1m_2/r^2\), from the chapter on Gravitational Force and Field. Both fall as \(1/r^2\) and act along the line joining the bodies. Two differences matter. Charge comes in two signs, so the electric force can repel as well as attract. And \(k\) is enormous compared with \(G\).
Set \(r = 30\,\text{cm}\) in the figure above: the force is \(0.60\,\text{N}\). At \(20\,\text{cm}\) it is \(1.35\,\text{N}\), which is \((30/20)^2 = 2.25\) times as much. Notice too that the two arrows stay equal in length, even though one charge is larger. Coulomb's law gives one number, the size of both forces.
(a) Find the force between \(2\,\mu\text{C}\) and \(3\,\mu\text{C}\) placed \(30\,\text{cm}\) apart in air. (b) Charges \(+4\,\mu\text{C}\) and \(-6\,\mu\text{C}\) attract with \(2.4\,\text{N}\). How far apart are they? (c) Two equal charges \(10\,\text{cm}\) apart repel with \(3.6\,\text{N}\). Find each charge.
(a) \(F\) \(= 9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}/(0.3)^2\) \(= 0.054/0.09\) \(= 0.6\,\text{N}\), repulsive because both charges are positive.
(b) \(r^2\) \(= k|q_1q_2|/F\) \(= 9 \times 10^9 \times 24 \times 10^{-12}/2.4\) \(= 0.09\,\text{m}^2\), so \(r = 0.3\,\text{m}\).
(c) \(q^2\) \(= Fr^2/k\) \(= 3.6 \times 0.01/(9 \times 10^9)\) \(= 4 \times 10^{-12}\,\text{C}^2\), so \(q = 2\,\mu\text{C}\).
Values used, the \(k\) route: \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
Two point charges attract each other with a force \(F\). Both charges are doubled and the distance between them is halved. What is the new force?
Two identical small metal spheres carry \(+6\,\mu\text{C}\) and \(-2\,\mu\text{C}\), \(30\,\text{cm}\) apart; they attract with \(1.2\,\text{N}\). They are touched together and put back \(30\,\text{cm}\) apart. Find the new force, in newtons.
A total charge of \(8\,\mu\text{C}\) is divided between two small balls placed \(20\,\text{cm}\) apart. Which division gives the largest force?
In 1785 the French engineer Charles-Augustin de Coulomb reported his measurements to the Academy of Sciences in Paris. He hung a light insulating rod from a very thin wire, with a small charged ball at one end. A second charged ball pushed the first away, and the wire twisted. The angle of twist measured the force. Halving the distance between the balls made the force about four times larger: the inverse square law.
Coulomb's law as a vector
A force has a direction, and the signs of the charges can supply it. Let \(q_1\) sit at position \(\vec r_1\) and \(q_2\) at \(\vec r_2\). The vector \(\vec r_2 - \vec r_1\) runs from \(q_1\) to \(q_2\). Divide it by its length \(r\) and you get the unit vector \(\hat r_{12}\), which points from \(q_1\) towards \(q_2\).
\[ \vec F_{21} = k\,\dfrac{q_1 q_2}{r^2}\,\hat r_{12} = k\,\dfrac{q_1 q_2}{|\vec r_2 - \vec r_1|^3}\,(\vec r_2 - \vec r_1) \]
Here \(\vec F_{21}\) is the force on \(q_2\) due to \(q_1\). Put the charges in with their signs. For like charges \(q_1q_2\) is positive, so \(\vec F_{21}\) points along \(\hat r_{12}\), away from \(q_1\). For unlike charges \(q_1q_2\) is negative, so the force points back towards \(q_1\). One formula gives both the size and the direction.
The force on \(q_1\) due to \(q_2\) comes from the same formula with the labels swapped. The unit vector then points from \(q_2\) to \(q_1\), the opposite way, so \(\vec F_{12} = -\vec F_{21}\). This is Newton's third law, from the chapter on Forces and Newton's Laws. The two forces are equal in size even when the charges are very different, as Fig 2.1 shows.
A charge \(q_A = +100\,\mu\text{C}\) sits at A \((1, 0, 2)\,\text{m}\) and \(q_B = +200\,\mu\text{C}\) at B \((4, 4, 2)\,\text{m}\). Find the size of the force between them, and write the force on each as a vector.
(a) \(\vec r_B - \vec r_A\) \(= 3\hat i + 4\hat j\), so \(r = 5\,\text{m}\) and \(\hat r_{AB}\) \(= 0.6\hat i + 0.8\hat j\).
(b) \(F\) \(= 9 \times 10^9 \times 100 \times 10^{-6} \times 200 \times 10^{-6}/25\) \(= 7.2\,\text{N}\).
(c) Both charges are positive, so B is pushed away from A: \(\vec F_B\) \(= 7.2(0.6\hat i + 0.8\hat j)\) \(= 4.32\hat i + 5.76\hat j\,\text{N}\).
(d) By the third law \(\vec F_A\) \(= -4.32\hat i - 5.76\hat j\,\text{N}\).
The \(k\) route, as in every point-charge example.
A charge \(q_1 = +2\,\mu\text{C}\) is at the origin and \(q_2 = -8\,\mu\text{C}\) is on the \(x\) axis at \(x = 0.3\,\text{m}\). The force on \(q_2\) is \(F_x\,\hat i\). Find \(F_x\) in newtons, with its sign.
A \(10\,\text{g}\) ball carrying \(1\,\mu\text{C}\) and a \(20\,\text{g}\) ball carrying \(4\,\mu\text{C}\) are \(60\,\text{cm}\) apart and free to move. Which statement is true at this instant?
Charges inside a medium
Put the same two charges, the same distance apart, into oil or water. The force between them becomes smaller. The molecules of the medium polarise, just as the paper did in the last lesson. Each molecule turns its opposite end towards each charge, and this partly screens the charges from each other.
The medium is described by its dielectric constant \(K\), a pure number. The permittivity of the medium is \(\varepsilon = K\varepsilon_0\), and it replaces \(\varepsilon_0\) in Coulomb's law. So the force falls by the factor \(K\). For air \(K\) is almost exactly 1. For water it is about 80, which is one reason water dissolves salt so well.
Two charges attract with \(0.8\,\text{N}\) in air. In a liquid of \(K = 4\) at the same distance, they attract with \(0.8/4 = 0.2\,\text{N}\). To get the air force back inside the liquid, the charges must come closer. \(Kr_{\text{medium}}^2\) \(= r_{\text{air}}^2\) gives \(r_{\text{medium}}\) \(= r_{\text{air}}/\sqrt K\). The other way round, two charges \(20\,\text{cm}\) apart in the liquid feel the force they would feel \(40\,\text{cm}\) apart in air.
Two charges are \(15\,\text{cm}\) apart in oil of dielectric constant \(K = 4\). At what distance in air would they feel the same force, in centimetres?
The electric force against gravity
In a hydrogen atom one electron is held by one proton, about \(5.3 \times 10^{-11}\,\text{m}\) away. Compare the two pulls between them. The electric pull is \(ke^2/r^2\) and the gravitational pull is \(Gm_em_p/r^2\). The distance cancels in their ratio, so the ratio is the same at every distance.
In a hydrogen atom the electron is \(5.3 \times 10^{-11}\,\text{m}\) from the proton. Find the electric force between them, and the ratio of the electric force to the gravitational force.
Both forces are attractive and act along the same line. Use \(F_e = ke^2/r^2\) and \(F_g = Gm_em_p/r^2\).
\(F_e\) \(= 9 \times 10^9 \times (1.6 \times 10^{-19})^2/(5.3 \times 10^{-11})^2\).
\(F_e\) \(= 2.30 \times 10^{-28}/2.81 \times 10^{-21}\) \(= 8.2 \times 10^{-8}\,\text{N}\). Also \(F_g\) \(= 3.6 \times 10^{-47}\,\text{N}\).
\(F_e/F_g\) \(= 8.2 \times 10^{-8}/3.6 \times 10^{-47}\) \(= 2.3 \times 10^{39}\). Gravity plays no part in holding an atom together.
Why then does gravity, not electricity, hold the Moon in its orbit? Large bodies carry almost exactly equal amounts of positive and negative charge. Their electric pushes and pulls cancel. Gravity has only one sign, so it never cancels. It adds up over every kilogram of a planet, and on that scale it wins.
Two small blocks, each of mass \(5\,\text{g}\) and charge \(1.0 \times 10^{-7}\,\text{C}\), rest on a table \(10\,\text{cm}\) apart. They are just about to slide apart. Take \(g = 10\,\text{m/s}^2\). Find the coefficient of friction.
Coulomb's law is for point charges, or for bodies small compared with their distance apart. Two large metal spheres close together are different. Their charges crowd towards or away from each other, so \(kq_1q_2/r^2\) with \(r\) between the centres is only rough.
Coal-fired power stations in India must clean fly ash from their smoke before it leaves the chimney. The smoke passes an electrostatic precipitator: thin wires at a high negative voltage give the dust particles a charge. Nearby earthed metal plates then pull the charged dust out of the gas stream. Well-run precipitators remove about 99 % of the ash, and the collected ash goes into cement and bricks.
- Using \(r\) in centimetres or \(q\) in microcoulombs inside \(kq_1q_2/r^2\).Convert first: charges in C, distances in m, so the force comes out in N.
- Thinking the larger charge feels the larger force.Both charges feel the same size of force, \(kq_1q_2/r^2\): a third-law pair.
- Forgetting the square: doubling \(r\) halves the force.Doubling \(r\) quarters the force, because \(F \propto 1/r^2\).
- Multiplying by \(K\) in a medium.The force in a medium is the air force divided by \(K\).