Coaching Centre
Electric fields
The electric field as force per unit charge (E = F/q), field direction and the test charge, rules for drawing field lines for point charges, dipoles and like charges, adding fields as vectors, and the uniform field between parallel plates (E = V/d).
A shark can find a flathead buried in the sand without seeing, smelling or hearing it. Tiny jelly-filled pores on its snout, the ampullae of Lorenzini, sense the electric field made by the fish's muscles and gills: as little as a few billionths of a volt per centimetre. Sharks are reading a field, the same idea physicists use to explain how charges push on each other without touching.
Force per unit charge
A charged object changes the space around it. Put another charge anywhere in that space and it feels a force. We call this region an electric field, and we measure its strength at a point by placing a small positive test charge q there and finding the force on it:
$$E = \frac{F}{q}$$
E is the force per coulomb, in newtons per coulomb (N C−1). It is a vector: its direction is the direction of the force on a positive charge. The test charge must be small so that it doesn't push the source charges around and change the field it is measuring.
Turn the equation round and you have the force on any charge in a known field, $F = qE$. For a positive charge the force is along the field; for a negative charge (such as an electron) it is opposite to the field.
This is the same idea as the gravitational field from Module 2, where $g = F/m$ is the force per kilogram. The difference is that gravity only ever attracts, while electric forces can attract or repel.
At a point in a thundercloud the electric field is 3.0 × 104 N C−1 directed downwards. Find the force on an electron at that point. ($q_e = -1.602 \times 10^{-19}$ C)
$F = qE = (1.602 \times 10^{-19})(3.0 \times 10^{4}) = 4.8 \times 10^{-15}$ N.
The electron is negative, so the force is opposite to the field: upwards.
Field lines
A field-line diagram shows the direction and relative strength of a field at a glance. The rules:
- Lines start on positive charges and end on negative charges (or at infinity).
- The arrow on a line shows the direction of the force on a positive test charge.
- The closer the lines, the stronger the field. The number of lines drawn from a charge is proportional to its size.
- Lines never cross, because the field can only have one direction at each point.
- Lines meet the surface of a conductor at right angles.
Around an isolated point charge the lines are radial and spread out with distance, so the field weakens. In 4C you'll see that it falls as the inverse square of distance:
$$E = \frac{kq}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}, \qquad k = 8.99 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$$
Adding fields
When there are several charges, the field at a point is the vector sum of the fields each charge would make on its own. Work out each field separately (size from $kq/r^2$, direction away from + or towards −), then add them as vectors.
A +4.0 nC charge and a −4.0 nC charge are 0.20 m apart. Find the electric field at the midpoint between them.
Each charge is 0.10 m from the midpoint: $E = \dfrac{kq}{r^2} = \dfrac{(8.99 \times 10^{9})(4.0 \times 10^{-9})}{(0.10)^2} = 3.6 \times 10^{3}$ N C−1.
The field of the positive charge points away from it, towards the negative charge. The field of the negative charge points towards it, which is the same direction.
$E = 3.6 \times 10^{3} + 3.6 \times 10^{3} = 7.2 \times 10^{3}$ N C−1, directed towards the negative charge. (For two equal like charges the two fields would cancel and E = 0 at the midpoint.)
The uniform field between parallel plates
Connect two flat, parallel metal plates to a power supply. Away from the edges, the field between them is uniform: the same size and direction everywhere, drawn as straight, parallel, evenly spaced lines from the positive plate to the negative plate. Its strength depends on the potential difference V across the plates and their separation d:
$$E = \frac{V}{d}$$
This gives E in volts per metre (V m−1), which is exactly the same unit as N C−1. You'll see why in 4D, where potential difference is defined. Doubling the voltage doubles the field; doubling the gap halves it. Year 12 Module 6 uses this field to steer beams of charged particles.
Charged ink droplets are steered between two plates 5.0 mm apart with 250 V across them. Find the field strength and the force on a droplet carrying a charge of −2.0 × 10−13 C.
$E = \dfrac{V}{d} = \dfrac{250}{5.0 \times 10^{-3}} = 5.0 \times 10^{4}$ V m−1, from the positive plate to the negative plate.
$F = qE = (2.0 \times 10^{-13})(5.0 \times 10^{4}) = 1.0 \times 10^{-8}$ N, towards the positive plate because the droplet is negative. The force is the same anywhere between the plates.
- Single +, 2.0 nC. P starts 5.0 cm away: E = 7.19 × 103 N C−1 at 0°. Predict E at 10.0 cm, then tap 5 grid squares further right to check: it should be a quarter, 1.80 × 103 N C−1.
- Note the force on the 1.0 nC test charge at 5.0 cm, 7.19 × 10−6 N. Show that F = qE.
- Dipole: P starts at the midpoint, 5 cm from each charge. The sim gives 1.44 × 104 N C−1 at 0° (towards the negative charge). Confirm it by adding the two fields.
- Like charges: the midpoint is a neutral point, E = 0. Turn on the vector grid and describe the field just above and below it.
- Parallel plates, 500 V: E = 5.00 × 103 V m−1 wherever P sits between the plates. Change the voltage and check E = V/d.
Test yourself
A charge of +2.0 μC at a point experiences a force of 0.60 N to the east. The electric field at that point is
E = F/q = 0.60/2.0 × 10−6 = 3.0 × 105 N C−1, in the direction of the force on a positive charge.
An electron is placed in a uniform field that points north. The electron
The force on a negative charge is opposite to the field, and a constant force gives a constant acceleration.
The potential difference between two parallel plates is halved and their separation is doubled. The field strength between them becomes
E = V/d: halving V halves E, and doubling d halves it again, so E is ¼ as large.
Which statement about electric field lines is correct?
Line density shows field strength. Lines never cross, point + to −, and show the force on a positive charge.
Two charges of +3.0 nC are 0.40 m apart. The electric field at the midpoint is
Each charge makes 674 N C−1 at the midpoint, but in opposite directions, so they cancel.
Label the field diagrams.
Which fields does each statement describe?
The electric field strength at a point is the force per unit on a small positive charge. Field lines start on charges, and the field is stronger where they are together. Between parallel plates the field is and its strength equals the potential difference divided by the .
Two small spheres carry charges of +5.0 nC and −5.0 nC and are 0.30 m apart. (a) Calculate the electric field at the midpoint. (b) Describe how the field would differ at the midpoint if both charges were +5.0 nC, and sketch or describe the field-line pattern near the midpoint in that case.
(a) Each charge is 0.15 m from the midpoint: E = kq/r² = (8.99 × 109)(5.0 × 10−9)/(0.15)² = 2.0 × 103 N C−1. The field of the positive charge points away from it and the field of the negative charge points towards it, so both point towards the negative charge and add: E = 4.0 × 103 N C−1 towards the negative charge.
(b) With two equal positive charges, the two fields at the midpoint are equal in size but opposite in direction, so they cancel and E = 0: a neutral point. Field lines leave both charges and curve away from each other, so no line passes through the midpoint; the lines from each charge bend outwards on either side of it.