Coulomb's Law
You rub a balloon on your hair, and it sticks to the wall. Nothing is glued, and nothing is pushing it. So what holds it there, and why does it fall off after a few minutes, or when you hold it farther from the wall? In this page you will learn the rule that gives the size of the electric force between charged objects. You will use it to answer these questions with numbers.
Charges push and pull
Matter has two kinds of electric charge, called positive and negative. A proton has a charge of $+1.60 \times 10^{-19}\ \text{C}$ and an electron has $-1.60 \times 10^{-19}\ \text{C}$. The unit is the coulomb (C). The size $e = 1.60 \times 10^{-19}\ \text{C}$ is the smallest amount of charge you can find on a free object.
The rule for the direction of the force is simple:
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Like charges repel (+ and +, or − and −).
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Opposite charges attract (+ and −).
The two charges always push or pull on each other with forces that are equal in size and opposite in direction. This is Newton's third law.
Why does the balloon stick to the wall? The charged balloon pulls on the charges inside the wall, so the wall's nearby surface becomes slightly oppositely charged. The balloon is then attracted to it. You will study that process in the next page, "Conservation of Charge: Induction, Conduction and Polarization."
Coulomb's law says $F = k\dfrac{|q_1 q_2|}{r^2}$. Two charged balloons repel each other. Without calculating, state what happens to the magnitude of the force if the…
Coulomb's law says $F = k\dfrac{|q_1 q_2|}{r^2}$. Two charged balloons repel each other. Without calculating, state what happens to the magnitude of the force if the distance between the centers of the balloons is doubled and the charges stay the same.
Coulomb's law
How strong is the force? Experiments by Charles-Augustin de Coulomb in the 1780s showed two things for small charged objects (point charges):
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The force gets bigger when either charge gets bigger. Double one charge and the force doubles.
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The force gets smaller quickly as the distance grows. Double the distance and the force becomes one fourth.
Two things to notice:
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The absolute value $|q_1 q_2|$ gives only the size of the force. You decide the direction with the rule above (like repel, opposite attract).
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The $r^2$ in the denominator is why the force drops so fast with distance. It is called an inverse-square law. Newton's law of gravity has the same shape, with masses instead of charges.
The coulomb is a huge amount of charge. Charges in everyday lab work are tiny, so you will see prefixes:
| Prefix | Symbol | Value |
|---|---|---|
| micro | μC | $10^{-6}\ \text{C}$ |
| nano | nC | $10^{-9}\ \text{C}$ |
Always convert to coulombs before you use $k$.
Two small charged spheres, $q_1 = 2.0 \times 10^{-6}\ \text{C}$ and $q_2 = 3.0 \times 10^{-6}\ \text{C}$, are placed $0.30\ \text{m}$ apart in air. Use…
Two small charged spheres, $q_1 = 2.0 \times 10^{-6}\ \text{C}$ and $q_2 = 3.0 \times 10^{-6}\ \text{C}$, are placed $0.30\ \text{m}$ apart in air. Use $k = 8.99 \times 10^{9}\ \text{N·m}^2/\text{C}^2$ to calculate the magnitude of the electric force between them. Write your answer in scientific notation. State whether the force is attractive or repulsive.
A proton and an electron in a hydrogen atom are separated by $5.3 \times 10^{-11}\ \text{m}$. Each has a charge of magnitude $1.60 \times 10^{-19}\ \text{C}$. Use…
A proton and an electron in a hydrogen atom are separated by $5.3 \times 10^{-11}\ \text{m}$. Each has a charge of magnitude $1.60 \times 10^{-19}\ \text{C}$. Use $k = 8.99 \times 10^{9}\ \text{N·m}^2/\text{C}^2$ to find the magnitude of the electric force between them. Give your answer in scientific notation.
Scaling: what happens if I change $q$ or $r$?
Often you do not need the actual number. You only need to know how the force changes. Use the structure of the formula: $F$ is proportional to $q_1$, proportional to $q_2$, and proportional to $1/r^2$.
| Change | Effect on $F$ |
|---|---|
| double one charge | $F$ doubles |
| triple one charge | $F$ is $3\times$ bigger |
| double both charges | $F$ is $4\times$ bigger |
| double the distance | $F$ is $\tfrac14$ as big |
| triple the distance | $F$ is $\tfrac19$ as big |
| halve the distance | $F$ is $4\times$ bigger |
You multiply the factors together when you change several things at once.
You can also test whether data follow an inverse-square law. If $r$ doubles and $F$ falls to $\tfrac14$, then $F\,r^2$ stays the same. This is how Coulomb, and your lab group, can check the law with measurements.
Two point charges exert a force of $12\ \text{N}$ on each other when they are $0.10\ \text{m}$ apart. Without changing the charges, Emma moves them to a separation of…
Two point charges exert a force of $12\ \text{N}$ on each other when they are $0.10\ \text{m}$ apart. Without changing the charges, Emma moves them to a separation of $0.40\ \text{m}$. Find the new force magnitude. Then find the force if, from the original setup, she instead triples one charge and keeps the distance at $0.10\ \text{m}$.
In a lab, Malik measures the force between two small charged spheres at different separations and records the data below. Both charges stay fixed. | Distance $r$ (m) |…
In a lab, Malik measures the force between two small charged spheres at different separations and records the data below. Both charges stay fixed.
| Distance $r$ (m) | Force $F$ (N) |
|---|---|
| 0.050 | 3.6 |
| 0.100 | 0.90 |
| 0.200 | 0.225 |
- a) Show that the data are consistent with an inverse-square relationship.
- b) Predict the force at $r = 0.025\ \text{m}$.
Force and motion: bringing in Newton's second law
The electric force is a force like any other. If it is the only force on a small object, then $F = ma$, so
$$a = \frac{F}{m}.$$A sphere with a mass of a few grams will accelerate very noticeably even with a tiny force, so convert grams to kilograms first.
The acceleration is only initial. As the spheres move apart (or together), $r$ changes, so $F$ and $a$ change too. That is why you cannot use the constant-acceleration equations for this motion.
Two identical small spheres each carry a charge of $+4.0 \times 10^{-7}\ \text{C}$ and are held $0.15\ \text{m}$ apart. Use…
Two identical small spheres each carry a charge of $+4.0 \times 10^{-7}\ \text{C}$ and are held $0.15\ \text{m}$ apart. Use $k = 8.99 \times 10^{9}\ \text{N·m}^2/\text{C}^2$.
- a) Calculate the force one sphere exerts on the other.
- b) Each sphere has a mass of $2.0 \times 10^{-3}\ \text{kg}$. Find the magnitude of the initial acceleration of one sphere when it is released.
Several charges on a line
When more than two charges are present, each pair interacts as if the others were not there. To find the net force on one charge:
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Find the force from each of the other charges, one at a time (use Coulomb's law for the size).
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Decide the direction of each force (like repel, opposite attract).
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Add them with signs. Choose $+x$ to the right and give each force a plus or minus sign.
A related question: can the net force on a third charge be zero? Yes. Between two charges of the same sign, the forces on a third charge point in opposite directions. Somewhere between them they cancel. That spot is closer to the smaller charge. The larger charge pushes harder at any given distance, so $q_3$ must be farther from the larger charge and nearer the smaller one for the two forces to be equal. Since $q_3$ appears in both forces, it cancels out of the equation, so the answer does not depend on $q_3$.
Three point charges lie on a straight line (the $x$-axis): $q_A = +2.0\ \mu\text{C}$ at $x = 0$, $q_B = -3.0\ \mu\text{C}$ at $x = 0.20\ \text{m}$, and…
Three point charges lie on a straight line (the $x$-axis): $q_A = +2.0\ \mu\text{C}$ at $x = 0$, $q_B = -3.0\ \mu\text{C}$ at $x = 0.20\ \text{m}$, and $q_C = +4.0\ \mu\text{C}$ at $x = 0.50\ \text{m}$. Use $k = 8.99 \times 10^{9}\ \text{N·m}^2/\text{C}^2$ and $1\ \mu\text{C} = 10^{-6}\ \text{C}$.
- a) Find the force exerted on $q_B$ by $q_A$ and the force exerted on $q_B$ by $q_C$, each with its direction.
- b) Find the net force on $q_B$, giving magnitude and direction.
- c) Explain why the net force on $q_B$ is not the same as the net force on $q_A$.
A charge $q_1 = +9.0\ \mu\text{C}$ is fixed at $x = 0$ and a charge $q_2 = +1.0\ \mu\text{C}$ is fixed at $x = 0.40\ \text{m}$. A small third charge $q_3$ is placed on…
A charge $q_1 = +9.0\ \mu\text{C}$ is fixed at $x = 0$ and a charge $q_2 = +1.0\ \mu\text{C}$ is fixed at $x = 0.40\ \text{m}$. A small third charge $q_3$ is placed on the line between them.
- a) Find the position $x$ where the net force on $q_3$ is zero. Show that your answer does not depend on the sign or size of $q_3$.
- b) Suppose $q_2$ were changed to $-1.0\ \mu\text{C}$. Explain, using the directions of the forces, whether there is still a point between the charges where the net force on $q_3$ is zero.
Back to the beginning: the balloon sticks to the wall because the charged balloon pulls on the opposite charges it induces in the wall. The force is real but tiny. If you hold the balloon farther from the wall, the force falls as $1/r^2$, and it soon cannot support the balloon's weight.