Field lines of point charges

What is it about?

A charge creates an electric field: any other charge feels a force there. The field strength is the force per test charge, $\vec E = \vec F / q_0$ , measured in N/C = V/m. For a point charge

E = k \cdot \frac{|Q|}{r^2}, \qquad k = \frac{1}{4\pi\varepsilon_0} \approx 8.99 \cdot 10^9\,\frac{\text{N}\,\text{m}^2}{\text{C}^2} .

The force on a charge $q$ is $F = q \cdot E$ : along the field lines for positive $q$ , opposite for negative $q$ (e.g. an electron). At the charge itself ( $r = 0$ ) the field is not defined. The fields of several charges add as vectors – the “vector addition” box shows the single fields head to tail and their sum in orange.

Field lines start at positive and end at negative charges; the field points along them, and dense lines mean a strong field (here 8 lines per nC). Equipotential lines (dashed green) join points of equal potential $\varphi = \sum k \cdot Q_i / r_i$ and are perpendicular to the field lines. Between the plates of a capacitor the field is uniform with $E = U / d$ .

The background colour shows the field strength on a logarithmic scale; the force arrow is capped for very strong fields (≫). Edge effects of the capacitor are neglected.

Try it

Drag the test charge around the dipole. Which way does the field point on the perpendicular bisector? Dipole
Single charge: double the distance from 10 cm to 20 cm. Single charge
Two equal charges: find the point where the field is zero. Two equal charges
Unequal charges: where do the extra lines go? +3 nC and −1 nC
Capacitor: does $E$ change between the plates? Capacitor

Tasks

Task 1: Point charge

$Q = 1.0\,\text{nC}$ . Find $E$ at $10\,\text{cm}$ and $20\,\text{cm}$ . Load task

Show solution

$E \approx 899\,\frac{\text{N}}{\text{C}}$ , at twice the distance a quarter: $\approx 225\,\frac{\text{N}}{\text{C}}$ . Show in the simulation

Task 2: Dipole

$\pm 1\,\text{nC}$ at $x = \mp 10\,\text{cm}$ . Find $E$ at the midpoint and the force on $q_0 = 1\,\text{nC}$ . Load task

Show solution

Both fields point towards the negative charge: $E \approx 2 \cdot 899 \approx 1798\,\frac{\text{N}}{\text{C}}$ , $F \approx 1.80\,\mu\text{N}$ . Show in the simulation

Task 3: Capacitor

$U = 1000\,\text{V}$ , $d = 10\,\text{cm}$ . Find $E$ and $F$ on $q_0 = 1\,\text{nC}$ . Load task

Show solution

$E = U / d = 10\,000\,\frac{\text{V}}{\text{m}}$ , $F = 10\,\mu\text{N}$ . Show in the simulation

Ideas for the classroom

Compare with field patterns from semolina in oil.
Common stumbling blocks:
- “Field lines are paths of charges.” – They show the force direction.
- “There is no field between the lines.” – The lines are only a selection.