The electric field and field lines

Hair that rises

Rub a balloon on your hair, then lift it slowly away. Your hair follows it upwards before the two touch. Something reaches across the gap and pulls.

In the last course we found the force between two charges with Coulomb's law. Now we ask a different question. Take away the second charge. Does the first one still do something to the space round it? Physicists say yes. The charge makes an electric field in the space around it, and any other charge that arrives there feels a force from that field.

What the field is

We need a way to measure the field at a point. Put a small positive test charge $q_0$ at that point and measure the force $F$ on it. The test charge must be so small that it does not push the other charges about. The field is the force on each unit of charge:

The unit is the newton for each coulomb (N/C). The field is a vector. Its direction is the direction of the force on a positive test charge.

The field is there whether or not a test charge is. A charge does not make the field when you bring it. It only feels the field that was already there. If you know the field $E$ at a place, the force on any charge $q$ placed there is

For a positive $q$ the force is along $E$. For a negative $q$ it is against $E$.

The field of a point charge

Put a test charge $q_0$ at a distance $r$ from a point charge $Q$. Coulomb's law gives the force $F = \dfrac{kQq_0}{r^2}$. Divide by $q_0$:

Here $k = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2$. The field points away from a positive $Q$ and towards a negative $Q$. It does not depend on the test charge at all. It falls as $1/r^2$: at twice the distance it is a quarter as large.

A worked example. A small ball carries $+2.5$ nC. What is its field $0.30$ m away? What force does a charge of $-3.0$ nC feel there?

1. The field. $E = \dfrac{kQ}{r^2} = \dfrac{9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times 2.5 \times 10^{-9}\ \text{C}}{(0.30\ \text{m})^2} = 250\ \text{N/C}$, directed away from the ball.
2. The force. $F = qE = 3.0 \times 10^{-9}\ \text{C} \times 250\ \text{N/C} = 7.5 \times 10^{-7}\ \text{N}$.
3. Its direction. The charge is negative, so the force is against the field. It points towards the ball.

At $0.60$ m the field is $62.5$ N/C, one quarter of $250$ N/C. That is the inverse square law at work.

Set the distance to $0.30$ m and then to $0.60$ m. By what factor does the field change? Now double the charge. What happens?

Several charges: add the fields

Fields add as vectors. This is the principle of superposition. At a point, find the field of each charge as if the others were absent. Then add the arrows, head to tail.

Two charges on a line. A charge of $+8.0$ nC and a charge of $-2.0$ nC are $0.40$ m apart. Find the field at the midpoint $M$, which is $0.20$ m from each.

1. The positive charge. $E_1 = \dfrac{9.0 \times 10^{9} \times 8.0 \times 10^{-9}}{(0.20)^2} = 1800\ \text{N/C}$. It points away from the $+8.0$ nC charge, which is towards the other charge.
2. The negative charge. $E_2 = \dfrac{9.0 \times 10^{9} \times 2.0 \times 10^{-9}}{(0.20)^2} = 450\ \text{N/C}$. It points towards the $-2.0$ nC charge, which is the same way.
3. The sum. Both point the same way, so $E = 1800 + 450 = 2250\ \text{N/C}$, towards the negative charge.

Between a positive and a negative charge the two fields point the same way, so they add. Between two charges of the same sign they point opposite ways, so they can cancel.

Fields at right angles. A charge of $+6.0$ nC sits on the $x$ axis at $x = 0.30$ m. A charge of $+8.0$ nC sits on the $y$ axis at $y = 0.40$ m. Find the field at the origin.

1. The first charge. $E_1 = \dfrac{9.0 \times 10^{9} \times 6.0 \times 10^{-9}}{(0.30)^2} = 600\ \text{N/C}$, pointing away from the charge, so along $-x$.
2. The second charge. $E_2 = \dfrac{9.0 \times 10^{9} \times 8.0 \times 10^{-9}}{(0.40)^2} = 450\ \text{N/C}$, along $-y$.
3. The sum. The two parts are at right angles, so $E = \sqrt{600^2 + 450^2}\ \text{N/C} = 750\ \text{N/C}$. Its direction makes an angle $\tan^{-1}(450/600) = 37^\circ$ with the $-x$ axis.

A point where the field is zero

Two like charges push on a test charge from opposite sides. Somewhere between them the pushes are equal. Take $+9.0$ nC and $+4.0$ nC, $0.50$ m apart. Let the point be a distance $x$ from the larger charge.

1. Equal fields. $\dfrac{kQ_1}{x^2} = \dfrac{kQ_2}{(d - x)^2}$, with $d = 0.50$ m.
2. Take the square root. $\dfrac{x}{d - x} = \sqrt{\dfrac{Q_1}{Q_2}} = \dfrac{3}{2}$. We want $x$, so $x = \dfrac{3}{5}d = 0.30\ \text{m}$.

The point is $0.30$ m from the $+9.0$ nC charge and $0.20$ m from the $+4.0$ nC charge. It is nearer the smaller charge.

For two charges of opposite sign, the fields point the same way between the charges, so there is no zero there. The zero lies outside, on the side of the smaller charge. For $+9.0$ nC and $-4.0$ nC, $0.50$ m apart, the same square-root step gives a point $1.0$ m beyond the $-4.0$ nC charge.

Start with equal positive charges. Where is the field zero? Now change one sign. Does the field still cross zero between the charges?

A charge in a field

A charge $q$ of mass $m$ in a field $E$ feels a force $qE$, so by Newton's second law its acceleration is

If the field is uniform, the same in size and direction everywhere, then $a$ is constant. The motion is as simple as a ball in the air, with $qE/m$ in place of $g$.

An electron in a field. An electron is in a field of $3.0 \times 10^{3}\ \text{N/C}$. Its charge is $1.6 \times 10^{-19}$ C and its mass is $9.1 \times 10^{-31}$ kg.

1. The force. $F = eE = 1.6 \times 10^{-19}\ \text{C} \times 3.0 \times 10^{3}\ \text{N/C} = 4.8 \times 10^{-16}\ \text{N}$.
2. The acceleration. $a = \dfrac{F}{m} = \dfrac{4.8 \times 10^{-16}\ \text{N}}{9.1 \times 10^{-31}\ \text{kg}} = 5.3 \times 10^{14}\ \text{m/s}^2$.
3. Compare with gravity. The weight of the electron is $mg = 9.1 \times 10^{-31}\ \text{kg} \times 9.8\ \text{m/s}^2 = 8.9 \times 10^{-30}\ \text{N}$. The electric force is $5.4 \times 10^{13}$ times larger. For small charged particles we leave out gravity.

The electron moves against the field, because its charge is negative.

Field lines

A field has a size and a direction at every point of space. How can we draw all that? Michael Faraday had a way. Draw lines so that the tangent to a line at any point is the direction of the field there. Put arrowheads on the lines.

The rules are short.

• Field lines start on positive charges and end on negative charges. If a charge is alone, the lines go off to a great distance, or come from one.
• The tangent to a line is the direction of the field. A positive charge placed on a line is pushed along it.
• Where the lines are close together the field is strong. Where they are far apart it is weak. The number of lines that leave a charge is in proportion to the size of the charge.
• Two field lines never cross. A crossing would give two directions for the field at one point, and the field has only one.
• Electrostatic field lines never form closed loops. A line that leaves a positive charge cannot come back to where it started, because it must end on a negative charge or go on to a great distance.

Look at the four drawings. In (a), the lines of a single charge spread out in all directions, so the field becomes weaker as the lines spread. In (b), the lines leave the positive charge and curve to the negative one. In (c), two like charges push their lines apart, and a gap appears between them where the field is zero. In (d), the lines between two plates are straight and evenly spaced. This is a uniform field: the same size and direction at every point.

Field lines are a way of drawing. They are not threads in space. The field is there between the lines too.

Practise with new numbers

Try it in Python

The program adds the field of two point charges at a point. It does this by adding the $x$ parts and the $y$ parts of each field. It uses the same two charges as the example at right angles.

Common mistakes

• Thinking the test charge makes the field. The field belongs to the source charges. The test charge only finds out what it is.
• Adding sizes when the directions differ. Fields add as vectors. Two fields of $600$ N/C and $450$ N/C at right angles give $750$ N/C, not $1050$ N/C.
• Forgetting the direction for a negative charge. The field of a negative charge points towards it. A negative charge in a field feels a force against the field.
• Forgetting to square the distance. The field of a point charge goes as $1/r^2$.
• Drawing field lines that cross, or that stop in empty space. A line starts on a positive charge and ends on a negative one.
• Mixing units. Change nC to C and cm to m before you use $k = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2$.

Remember these

• $E = F/q_0$, in N/C. The force on a charge $q$ is $F = qE$.
• A point charge: $E = \dfrac{kQ}{r^2}$, away from $+Q$ and towards $-Q$.
• Fields of several charges add as vectors (superposition).
• A charge in a field accelerates at $a = \dfrac{qE}{m}$. A negative charge goes against the field.
• Field lines start on $+$ and end on $-$. They never cross. Close lines mean a strong field. Parallel, evenly spaced lines mean a uniform field.

Check yourself

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

1. A test charge of $+2.0$ nC feels a force of $6.0 \times 10^{-6}$ N at a point. What is the electric field there?

2. The field of a point charge is $400$ N/C at a distance of $1.0$ m. What is it at $4.0$ m?

3. Can the electric field at a point be zero while a charge sits at that point? Can it be zero at a point midway between a positive and a negative charge of equal size?

4. Why can two electric field lines never cross?