Why is potential zero at infinity?

It is a convenient convention: a charge infinitely far away has no interaction energy with the field source.

Can potential be negative?

Yes, near a negative charge. Work would be done by the field bringing a positive charge in from infinity.

Why use potential instead of field strength?

Potential is a scalar, so contributions from several charges add as plain numbers. Energy calculations also follow directly from W = Q V.

Why are equipotentials perpendicular to field lines?

Moving along an equipotential involves no change in potential and so no work. That requires the force, and therefore the field, to have no component along it.

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112, 114–115ElectricityStretch

Electric Potential

Field strength describes the force on a charge. $\textbf{Electric potential}$ describes the energy: the work done per unit charge bringing a positive charge from infinity to a point. Because potential is a scalar, it is often far easier to work with than the field.

35 min Topic 7 · Electric and Magnetic Fields

What you'll be able to do

Define electric potential and potential difference in terms of work
Use $V = kQ/r$ for a point charge and add potentials as scalars
Relate field strength to potential gradient, $E = -\Delta V/\Delta r$
Describe equipotential surfaces and explain why they are perpendicular to field lines

Potential and potential energy

The electric potential at a point is the work done per unit positive charge in bringing a charge from infinity to that point. It is measured in volts, where $1\ \mathrm{V} = 1\ \mathrm{J\,C^{-1}}$ , and infinity is defined as zero potential.

A charge $Q$ at a point of potential $V$ has electric potential energy $E_{p} = QV$ . The work done moving a charge between two points is the charge times the potential difference.

Near a positive charge the potential is positive, since work must be done to push a positive charge closer. Near a negative charge it is negative.

V = \dfrac{W}{Q}

Potential, work per unit charge.

W = Q\Delta V

Work moving a charge through a p.d.

Tip — Potential is a property of a point in the field; potential energy belongs to a charge placed there. Keep the two apart.

Potential of a point charge

For a point charge, $V = kQ/r$ . It falls as $1/r$ , not $1/r^{2}$ — more gradually than field strength.

Because potential is a $\textbf{scalar}$ , the potential at a point due to several charges is simply the algebraic sum of each one. Signs matter, but directions do not.

Two charges produce potential energy $E_{p} = \dfrac{kQ_{1}Q_{2}}{r}$ , the work needed to assemble them from infinite separation.

V = \dfrac{kQ}{r}

Potential of a point charge.

E_{p} = \dfrac{kQ_{1}Q_{2}}{r}

Potential energy of a pair.

Example 1Charges of

+2.0\ \mathrm{nC}

and

-6.0\ \mathrm{nC}

are 0.40 m apart. Find the potential at the midpoint.

1Each is 0.20 m from the midpoint.

V = \dfrac{8.99\times10^{9}}{0.20}\left(2.0 - 6.0\right)\times10^{-9}

= 44.95 \times (-4.0)

Answer

-180\ \mathrm{V}

Field as potential gradient

Field strength and potential are linked by $E = -\dfrac{\Delta V}{\Delta r}$ : the field is the negative of the potential gradient. The minus sign means the field points from high potential to low — the direction a positive charge would naturally move.

For a uniform field this is just $E = V/d$ again, since potential falls steadily between the plates.

On a graph of $V$ against $r$ , the field strength at any point is the negative of the gradient. Where the graph is steepest, the field is strongest.

E = -\dfrac{\Delta V}{\Delta r}

Field is the negative potential gradient.

Example 1Potential falls from 300 V to 120 V over 0.060 m in a uniform field. Find the field strength.

\Delta V = 120 - 300 = -180

E = -\dfrac{-180}{0.060}

Answer

3.0\times10^{3}\ \mathrm{V\,m^{-1}}

, pointing towards lower potential.

Tip — The field points "downhill" in potential. A positive charge released at rest always moves to lower potential; an electron moves to higher.

Equipotentials

An $\textbf{equipotential}$ is a line or surface joining points of equal potential. Moving a charge along one requires no work, because $\Delta V = 0$ .

Equipotentials are always $\textbf{perpendicular}$ to field lines. If they were not, the field would have a component along the equipotential and work would be done moving along it — a contradiction.

For a point charge, equipotentials are concentric spheres, getting further apart with distance since $V$ falls as $1/r$ . For parallel plates they are equally spaced planes parallel to the plates.

Equation recap

V = W/Q

Definition.

V = kQ/r

Point charge.

E_{p} = kQ_{1}Q_{2}/r

Potential energy of a pair.

E = -\Delta V/\Delta r

Field from potential.

Common mistakes to avoid

Using

V = kQ/r^{2}

Potential falls as

1/r

; field strength falls as

1/r^{2}

Adding potentials as vectors.

Potential is a scalar; add algebraically with signs.

Saying work is done moving along an equipotential.

No potential difference means no work.

Treating potential and potential energy as the same.

Potential energy is

QV

, belonging to a specific charge.

Key takeaways

Potential is work per unit positive charge from infinity, in volts.
For a point charge $V = kQ/r$ ; potentials add as scalars.
Potential energy of a charge is $QV$ ; of a pair, $kQ_{1}Q_{2}/r$ .
Field strength is the negative potential gradient.
Equipotentials are perpendicular to field lines, and no work is done along them.

Test yourself

Answer a few questions, then review anything you missed.

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