Forces between several charges

A crowd of charges

Real charges are never alone. A charged ball on a bench has a table, a hand and a wall nearby, all with charges. Does the force of one charge change when another charge comes near?

Experiments say that it does not. The force that charge $q_1$ exerts on $q_3$ is the same whether $q_2$ is there or not. This is the principle of superposition. It gives us a method:

1. Find the force on the charge from each of the others, as if the others were not there.
2. Add the forces as vectors.

The only hard part is that forces have directions. Let us start with the easy case.

Charges on a straight line

When all the charges are on one line, every force points along the line. Choose one direction as positive, and add the forces with their signs.

A worked example. Three charges lie on a line: $A = +6.0\ \mu\text{C}$ at $x = 0$, $B = -3.0\ \mu\text{C}$ at $x = 0.15$ m, and $C = +4.0\ \mu\text{C}$ at $x = 0.35$ m. Find the net force on $B$.

1. The force from A. The distance is $0.15$ m. $F_{AB} = \dfrac{kq_Aq_B}{r^2}$. The top is $9.0 \times 10^9 \times 6.0 \times 10^{-6} \times 3.0 \times 10^{-6}\ \text{N m}^2 = 0.162\ \text{N m}^2$ and the bottom is $(0.15\ \text{m})^2 = 0.0225\ \text{m}^2$. So $F_{AB} = 7.2\ \text{N}$. $A$ is positive and $B$ is negative, so they attract. The force on $B$ points towards $A$, which is the $-x$ direction. So $F_{AB} = -7.2\ \text{N}$.
2. The force from C. The distance is $0.35\ \text{m} - 0.15\ \text{m} = 0.20\ \text{m}$. $F_{CB} = \dfrac{kq_Cq_B}{r^2}$. The top is $9.0 \times 10^9 \times 4.0 \times 10^{-6} \times 3.0 \times 10^{-6}\ \text{N m}^2 = 0.108\ \text{N m}^2$ and the bottom is $(0.20\ \text{m})^2 = 0.040\ \text{m}^2$. So $F_{CB} = 2.7\ \text{N}$. $C$ is positive, so it also attracts $B$, towards $+x$. So $F_{CB} = +2.7\ \text{N}$.
3. The sum. $F = -7.2\ \text{N} + 2.7\ \text{N} = -4.5\ \text{N}$. The net force on $B$ is $4.5$ N towards $A$.

What if $B$ were $+3.0\ \mu\text{C}$? Then both forces change direction, and the net force is $4.5$ N towards $C$.

Where is the force zero?

Two positive charges $q_1$ and $q_2$ are fixed a distance $d$ apart. Where could a small positive charge be placed so that the net force on it is zero?

Between the two charges, the forces point in opposite directions, so they can cancel. Outside, both forces point the same way, so they never cancel. Let the point be at a distance $x$ from $q_1$. The two forces are equal when

The constant $k$ and the test charge $q_3$ cancel. Then $\dfrac{q_1}{x^2} = \dfrac{q_2}{(d-x)^2}$. Take the square root of both sides:

The point of zero force is nearer the smaller charge. The bigger charge pushes harder from the same distance, so we must go farther from it.

An example. The charges are $4\ \mu\text{C}$ and $9\ \mu\text{C}$, and they are $0.60$ m apart. Then $\dfrac{x}{0.60\ \text{m} - x} = \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$. So $3x = 2(0.60\ \text{m} - x)$, which gives $5x = 1.20\ \text{m}$ and $x = 0.24$ m from the $4\ \mu\text{C}$ charge. It is $0.36$ m from the $9\ \mu\text{C}$ charge. As we expect, it lies nearer the smaller charge.

For two unlike charges, the point of zero force is outside the line between them. In the gap, both forces point the same way. Outside, on the side of the smaller charge, the forces point opposite ways, and the nearer smaller charge can match the farther bigger one.

Move the test charge until the net force is zero. Then make the right charge $9\ \mu\text{C}$ and the left charge $1\ \mu\text{C}$. Which way does the point move?

Charges that are not on a line

Now the forces point in different directions. We add them as vectors. There are two good methods.

• Components. Split each force into an $x$ part and a $y$ part. Add all the $x$ parts. Add all the $y$ parts. Then the net force is $F = \sqrt{F_x^2 + F_y^2}$, at an angle $\theta$ with $\tan\theta = F_y/F_x$.
• The parallelogram. Draw the two forces from one point, and complete the parallelogram. The diagonal is the sum.

A worked example: the equilateral triangle. Three equal charges of $+3.0\ \mu\text{C}$ sit at the corners of an equilateral triangle of side $0.20$ m. Find the force on one of them.

1. One neighbour. $F = \dfrac{kq^2}{a^2} = \dfrac{9.0 \times 10^9 \times (3.0 \times 10^{-6})^2}{(0.20)^2}\ \text{N} = 2.0\ \text{N}$. This push is directed away from that neighbour.
2. The angle. The two pushes make an angle of $60^\circ$ with each other. The picture shows them.
3. The sum. By symmetry, the net force lies along the line that bisects the angle, so each push contributes $F\cos 30^\circ$ along it. $F_{\text{net}} = 2F\cos 30^\circ = \sqrt{3}\,F = 3.5\ \text{N}$.

The net force points away from the middle of the opposite side.

A worked example: a right angle. A charge of $+2.0\ \mu\text{C}$ sits at the origin. A charge of $+5.0\ \mu\text{C}$ is at $(0.25\ \text{m}, 0)$, and a charge of $-3.0\ \mu\text{C}$ is at $(0, 0.35\ \text{m})$. Find the net force on the $2.0\ \mu\text{C}$ charge.

1. From the $+5.0\ \mu\text{C}$ charge. $F_1 = \dfrac{kqq_1}{r^2}$. The top is $9.0 \times 10^9 \times 2.0 \times 10^{-6} \times 5.0 \times 10^{-6}\ \text{N m}^2 = 0.090\ \text{N m}^2$ and the bottom is $(0.25\ \text{m})^2 = 0.0625\ \text{m}^2$. So $F_1 = 1.44\ \text{N}$. It is a push, away from that charge. So it points along $-x$: $F_x = -1.44\ \text{N}$.
2. From the $-3.0\ \mu\text{C}$ charge. $F_2 = \dfrac{kqq_2}{r^2}$. The top is $9.0 \times 10^9 \times 2.0 \times 10^{-6} \times 3.0 \times 10^{-6}\ \text{N m}^2 = 0.054\ \text{N m}^2$ and the bottom is $(0.35\ \text{m})^2 = 0.1225\ \text{m}^2$. So $F_2 = 0.44\ \text{N}$. It is a pull, towards that charge. So it points along $+y$: $F_y = +0.44\ \text{N}$.
3. The sum. The two forces are at a right angle, so $F = \sqrt{(1.44\ \text{N})^2 + (0.44\ \text{N})^2} = 1.51\ \text{N}$. The angle is $\tan\theta = \dfrac{0.44}{-1.44}$, which puts the force $17^\circ$ above the $-x$ direction.

Make $q_1$ negative. Which arrow flips? Now find values of $q_1$ and $q_2$ for which the net force points straight up.

Try it in Python

The program has a function that finds the net force on a charge from any list of other charges. It works with the signs and the directions for you. It repeats the right-angle example above.

Common mistakes

• Adding the sizes of the forces. Forces are vectors. Add them with their directions.
• Thinking that a third charge changes the force between two others. It does not. It only adds a force of its own.
• Putting the point of zero force in the middle. It is nearer the smaller charge, unless the charges are equal.
• Mixing up the directions. A push goes away from the other charge, and a pull goes towards it. Draw an arrow for every force before you add.
• Taking the wrong distance. For a charge between two others, use the distance to each of them, not the whole gap.
• Using the wrong angle. In the equilateral triangle the two pushes make $60^\circ$ with each other. Each one makes $30^\circ$ with the net force.

Remember these

• Superposition: $\vec F = \vec F_1 + \vec F_2 + \dots$. The force between two charges does not change when other charges are present.
• On a line: choose a positive direction, and add with signs.
• In a plane: add components, then $F = \sqrt{F_x^2 + F_y^2}$.
• Two equal charges $q$ at an angle of $60^\circ$ to each other give a net force $\sqrt{3}\,F$.
• Between two like charges, the force on a third is zero where $\dfrac{x}{d - x} = \sqrt{\dfrac{q_1}{q_2}}$, nearer the smaller charge.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A charge $+q$ sits exactly midway between two other charges, each $+Q$. What is the net force on the charge $+q$?

2. Charges of $9\ \mu\text{C}$ and $4\ \mu\text{C}$ are fixed $50$ cm apart. Where between them would a small positive charge feel no force?

3. Two other charges push a third charge with forces of $6$ N and $8$ N. The two forces are at a right angle. What is the net force?

4. A student says: "When I bring charge $C$ near charges $A$ and $B$, the force between $A$ and $B$ becomes smaller." Is she right?