EG1216 · Physics 2 — Electricity & Magnetism
Theme 1 · Electrostatics — charge & the electric field

The electric field & field lines

Give a single charge a "reach" into the space around it — so you can find the force on any charge placed there, before it is even there.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §23.4, §23.6.

Before you start

What you need first

  • Topic 1 — charge — the two signs; the elementary charge.
  • Topic 2 — Coulomb's law — \(F = k|q_1q_2|/r^2\), \(k=8.99\times10^{9}\), and adding forces as arrows (superposition).

What you'll be able to do

  • Define the electric field as force per unit charge, \(E = F/q\).
  • Find the field of a point charge, \(E = k|q|/r^2\), and its direction.
  • Find the force on a charge sitting in a field, \(F = qE\).
  • Read and draw field lines.
  • Add the fields of several charges (superposition).

The idea

What is an electric field?

A charge can push another charge without touching it. How? It changes the space all around itself. We call that changed space the electric field, symbol \(E\): at every point it is "ready to push" any charge placed there.

Think of a heater 🔥. Stand near it and you feel a lot of warmth; stand far and you feel less. The warmth fills the space around the heater even when nobody is standing there. A charge fills the space around it with a push in exactly the same way.

A charge makes a field everywhere around it. Drop a small test charge in, and it feels a force — stronger close in, weaker far away.
Q (the source) test charge
\(Q\) fills the space with a field; a charge placed anywhere feels a push.

The equation

Definition: field = force per charge

To measure the field at a point, put a small test charge \(q\) there, measure the force \(F\) it feels, and divide:

$$ E = \frac{F}{q} $$
where:
SymbolMeaningSI unit
Eelectric field strength — the force each coulomb of charge would feel hereN/C
Fforce on the test charge placed at the pointN
qthe test charge you placed thereC
\(E\) does not depend on the test charge — it is a property of the space, set by whatever charges are around. The test charge just lets us measure it.

The field made by a single point charge

Combine Coulomb's law with \(E = F/q\) and the test charge cancels, leaving the field a distance \(r\) from a charge \(q\):

$$ E = k\,\frac{|q|}{r^{2}} $$
where:
SymbolMeaningSI unit
Efield strength a distance \(r\) from the chargeN/C
qthe charge making the field (use its size)C
rdistance from the charge to the pointm
kCoulomb's constant \(=8.99\times10^{9}\)N·m²/C²
Graph of field strength falling off as one over r squared with distance
Same inverse-square shape as Coulomb's law: the field falls off as \(1/r^2\).

Which way does the field point?

The field points the way a tiny positive test charge would be pushed:

  • Around a positive charge, the field points away (outward).
  • Around a negative charge, the field points toward it (inward).
Field lines pointing outward from a positive charge
Out of a positive charge.
Field lines pointing inward toward a negative charge
Into a negative charge.
Memory aid: the field arrow shows the push on a + charge — out of +, into −.
📐 Worked example 1

Field from a point charge

Find the electric field a distance \(r = 0.30\) m from a charge \(q = +5\ \mu\text{C}\).

1Use the point-charge field with the size of \(q\):
$$ E = (8.99\times10^{9})\,\frac{5\times10^{-6}}{(0.30)^{2}} $$
2Work it out (bottom \(=0.09\)):
$$ E \approx 5.0\times10^{5}\ \text{N/C} $$
Because \(q\) is positive, the field points away from the charge at that point.

The force a field puts on a charge

Turn the definition around: if you already know the field \(E\) at a point, the force on a charge \(q\) placed there is

$$ F = q\,E $$
where:
SymbolMeaningSI unit
Fforce on the charge sitting in the fieldN
qthe charge placed in the fieldC
Ethe field at that spotN/C
A + charge is pushed along the field; a − charge is pushed against it. (\(E=F/q\) and \(F=qE\) are the same equation rearranged — pick the one with your unknown.)

Drawing fields: field lines

We picture a field with field lines — smooth lines whose arrow shows which way a + charge is pushed. The rules:

  • Lines start on + charges and end on − charges.
  • Where lines are closer together, the field is stronger.
  • Lines never cross (the field has one direction at each point).
  • The arrow is the field's direction; a line's tangent gives \(E\) there.
Field lines turn an invisible field into a picture: direction from the arrows, strength from how crowded the lines are.
Field lines running from a positive charge to a nearby negative charge
A dipole: lines leave the + charge and arrive on the − charge.

Several charges? Add the fields

Just like forces, the field from several charges is the sum of each charge's field, added as arrows (with direction):

1Find \(E_1\) from the first charge (size and direction).
2Find \(E_2\) from the second charge.
3Add them with direction → the total field.
Same fields, same way → they add. Opposite ways → they subtract. It is exactly the superposition you used for forces in Topic 2.
📐 Worked example 2

Field from two charges (a twist)

Charges \(q_1 = +2\ \mu\text{C}\) and \(q_2 = -2\ \mu\text{C}\) are \(0.20\) m apart. Find the field at the midpoint \(P\) (0.10 m from each).

At \(P\), the field from \(+q_1\) points away from it (to the right); the field from \(-q_2\) points toward it (also to the right). Same direction → they add.

q₁ + q₂ − P E₁, E₂
At the midpoint both fields point right, so they add.
1Field from \(q_1\) at \(P\) (points right, away from +):
$$ E_1 = (8.99\times10^{9})\,\frac{2\times10^{-6}}{(0.10)^{2}} \approx 1.8\times10^{6}\ \text{N/C} $$
2Field from \(q_2\) at \(P\) (points right, toward −) — same size:
$$ E_2 \approx 1.8\times10^{6}\ \text{N/C} $$
3Same direction → add:
$$ E_{\text{total}} \approx 3.6\times10^{6}\ \text{N/C to the right} $$
Between a + and a − charge, the two fields reinforce — that is why the field is strong in the gap of a dipole.

✏️ Try it yourself

A charge \(q = -8\ \mu\text{C}\) sits alone.

(a) Find the field at a point \(r = 0.10\) m away, and its direction.
(b) A small \(+2\ \text{nC}\) test charge is then placed at that point. Find the force on it.

(a) Step 1. \(E = (8.99\times10^{9})\dfrac{8\times10^{-6}}{(0.10)^{2}}\) Step 2. \(E \approx 7.2\times10^{6}\ \text{N/C}\), pointing toward the charge (it is negative). (b) \(F = qE = (2\times10^{-9})(7.2\times10^{6}) \approx 1.4\times10^{-2}\ \text{N}\) — about \(0.014\) N, pulling the + test charge toward \(q\).

Common mistakes

MistakeFix
Mixing up \(E = F/q\) and \(F = qE\).Same equation rearranged. Pick the one with your unknown.
Using \(r\) instead of \(r^2\) in \(E = k|q|/r^2\).The distance is squared.
Forgetting the field has a direction.Out of +, into −. Add fields as arrows.
Wrong units for \(E\).\(E\) is in N/C (newtons per coulomb).

Recap — the whole topic on one screen

IdeaWhat you own now
FieldThe "reach" of a charge into space; \(E = F/q\), in N/C.
Point charge\(E = k\,\dfrac{|q|}{r^2}\) — same inverse-square shape as Coulomb's law.
DirectionOut of a + charge, into a − charge.
Force from a field\(F = qE\); + goes along \(E\), − goes against it.
Field linesStart on +, end on −, never cross; crowded = strong.
Many chargesSuperposition — add each field as an arrow.

Next topic

The field of continuous charge distributions

So far our charges are single points. Real objects are spread out — rods, rings, and sheets of charge. Next we add up the field of a smooth distribution using an integral.

→ Topic 4 · Field of continuous distributions