EG1216 · Physics 2 — Electricity & Magnetism

Theme 2 · Electric potential & capacitance

Electric potential & potential difference

Instead of tracking force and direction everywhere, track one number per point: the energy each coulomb would have. That number is voltage.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §25.1–25.2.

Before you start

What you need first

Topic 5 — work and energy, and the uniform field between parallel plates.
Topic 3 — the force on a charge, $F = qE$.
Energy in joules (Physics 1).

What you'll be able to do

Say what electric potential $V$ (voltage) means: energy per charge.
Use $U = qV$ for the energy of a charge, and $W = q\,\Delta V$ for the work to move it.
Use the volt, $1\ \text{V} = 1\ \text{J/C}$.
Link voltage to a uniform field with $V = Ed$.

The idea

Potential is "electric height"

Electric potential $V$ (also called voltage) is the electric energy that each coulomb of charge would have at a point. It is energy per charge — a single number at every point, with no direction.

Picture it as height. Near a $+$ charge the potential is high, like the top of a hill; far away it is low. A $+$ charge "rolls downhill" — from high $V$ to low $V$ — which is exactly the way the field pushes it. The big win: because $V$ is just a number, it is far easier to work with than the field's arrows.

The energy of a charge at a potential

If a charge $q$ sits at a point where the potential is $V$, its electric potential energy is simply charge times potential:

$$ U = qV $$

where:
Symbol	Meaning	SI unit
U	electric potential energy of the charge	J
q	the charge (keep its sign)	C
V	potential at that point	V = J/C

This is exactly why the volt is a joule per coulomb: multiply a voltage by a charge and you get an energy.

Potential difference (what a battery gives)

Potential difference $\Delta V$ is the difference in voltage between two points. This is what a battery provides — a "9 V battery" means $\Delta V = 9$ V between its terminals.

Like a height difference between two steps on a staircase: only the difference matters for the energy, not the height you measure from. ($\Delta$, "delta," just means "the change in.")

The work to move a charge

Moving a charge $q$ through a potential difference $\Delta V$ changes its energy by

$$ W = q\,\Delta V $$

where:
Symbol	Meaning	SI unit
W	work (energy) to move the charge	J
q	the charge being moved	C
ΔV	potential difference it moves through	V

Move more charge, or move it through a bigger voltage → more energy.

📐 Worked example 1

Energy and work from voltage

(a) A charge $q = 2.0\ \mu\text{C}$ sits where the potential is $V = 500$ V — find its energy. (b) How much work moves a charge $q = 3.0\ \mu\text{C}$ through $\Delta V = 200$ V?

1Energy from $U = qV$:

$$ U = (2.0\times10^{-6})(500) = 1.0\times10^{-3}\ \text{J} $$

2Work from $W = q\,\Delta V$:

$$ W = (3.0\times10^{-6})(200) = 6.0\times10^{-4}\ \text{J} $$

The bridge to the field

Voltage and a uniform field

Between two parallel plates the field is uniform, and voltage and field are linked very simply. The voltage across a gap $d$ is the field times the distance:

$$ V = Ed \qquad E = \frac{V}{d} $$

This is why a field can also be measured in volts per metre (V/m) — the same unit as N/C.

A uniform field $E$ across a gap $d$; the dashed lines are equipotentials. The voltage across the gap is $V = Ed$.

📐 Worked example 2

Voltage across plates, and field from a battery

(a) A uniform field $E = 1000$ V/m sits across a gap $d = 0.050$ m — find the voltage. (b) A 12 V battery is connected across plates $d = 0.020$ m apart — find the field.

1Voltage from $V = Ed$:

$$ V = (1000)(0.050) = 50\ \text{V} $$

2Field from $E = V/d$:

$$ E = \frac{12}{0.020} = 600\ \text{V/m} $$

✏️ Try it yourself

A 9.0 V battery is connected across two parallel plates $d = 0.030$ m apart.

(a) Find the field between the plates.
(b) Find the work needed to move a charge $q = 4.0\ \mu\text{C}$ from one plate to the other.

(a) $E = \dfrac{V}{d} = \dfrac{9.0}{0.030} = 300\ \text{V/m}.$ (b) The charge crosses the full $\Delta V = 9.0$ V, so $W = q\,\Delta V = (4.0\times10^{-6})(9.0) = 3.6\times10^{-5}\ \text{J}.$

Common mistakes

Mistake	Fix
Treating voltage as a vector.	$V$ is a scalar — a number with a sign, no direction.
Dropping the sign of the charge in $U=qV$.	A negative charge at a positive potential has negative energy.
Confusing $V$ (a point's voltage) with $\Delta V$ (a difference).	Energy to move a charge uses the difference $\Delta V$ between start and end.
Forgetting $E$ can be in V/m.	V/m and N/C are the same unit; $V = Ed$ shows why.

Recap — the whole topic on one screen

Idea	What you own now
Potential	$V$ = energy per charge (a scalar); "electric height."
Energy	$U = qV$; the volt is $1\ \text{J/C}$.
Work	$W = q\,\Delta V$ to move a charge through a potential difference.
Uniform field	$V = Ed$, $E = V/d$; field in V/m.

Next topic

Potential of point charges, and E from V

Next we find the actual voltage around a charge, $V = kQ/r$, add the voltages of several charges (just numbers!), and turn the link around to get the field back from how the voltage changes.

→ Topic 10 · Potential & PE of point charges; E from V

Mistake	Fix
Treating voltage as a vector.	\(V\) is a scalar — a number with a sign, no direction.
Dropping the sign of the charge in \(U=qV\).	A negative charge at a positive potential has negative energy.
Confusing \(V\) (a point's voltage) with \(\Delta V\) (a difference).	Energy to move a charge uses the difference \(\Delta V\) between start and end.
Forgetting \(E\) can be in V/m.	V/m and N/C are the same unit; \(V = Ed\) shows why.

Idea	What you own now
Potential	\(V\) = energy per charge (a scalar); "electric height."
Energy	\(U = qV\); the volt is \(1\ \text{J/C}\).
Work	\(W = q\,\Delta V\) to move a charge through a potential difference.
Uniform field	\(V = Ed\), \(E = V/d\); field in V/m.