Electric potential and energy

What you'll learn

What electric potential means and how it links to work done and energy.
How to use $ΔEp=qΔV\Delta E_p = q\Delta V$ for charges moving through potential differences.
How potential behaves in uniform fields and around point charges.
How to handle signs, electronvolts, equipotentials and superposition.

The starting point: fields, work and charge

An electric field is a region where a charge experiences a force. Electric field strength, symbol $E$ , is force per unit positive charge:

\frac{F}{q}

Its units are newtons per coulomb, N C^-1. Later in this topic you will also meet the equivalent unit volts per metre, V m^-1.

A test charge is a very small charge used to investigate a field without noticeably changing the field itself. By convention, field direction is the direction of the force on a positive test charge.

Work done is energy transferred by a force. In electric fields, the important idea is not just “how large is the force?”, but “how much energy is transferred per coulomb of charge?”.

Key Idea

Potential is an energy idea

Electric potential lets you solve many electric-field problems using energy per charge instead of force and distance.

Electric potential

Definition

Electric potential

The electric potential at a point, symbol $V$ , is the work done per unit positive charge in bringing a small positive test charge from infinity to that point, without changing its kinetic energy.

In equation form:

\frac{W}{q}

The unit is the volt, V. One volt is one joule per coulomb, J C^-1.

For isolated point charges, we usually choose the electric potential to be zero at infinity. This is similar to gravitational potential being defined relative to infinity in fields around planets and stars.

Potential is scalar

Electric potential is a scalar quantity. This means it has magnitude and sign, but no direction.

That is different from electric field strength, which is a vector. Electric field strength has both magnitude and direction.

Potential difference and energy changes

Definition

Potential difference

The potential difference between two points is the work done per unit charge in moving a charge between those points.

If a charge $q$ moves through a potential difference $ΔV\Delta V$ , its change in electric potential energy is:

ΔEp=qΔV\Delta E_p = q\Delta V

This is one of the most important equations in this topic.

The work done by the electric field is the negative of the change in potential energy:

Wfield=−qΔVW_{\text{field}} = -q\Delta V

So if electric potential energy decreases, kinetic energy can increase.

Example

Energy change across a potential difference

An ion with charge $+3.0 \times 10^{-6}$ C moves from a point at 12 V to a point at 45 V. Find the change in electric potential energy and the work done by the electric field.

Calculate the potential difference: $ΔV=VB−VA=45−12=33\Delta V = V_B - V_A = 45 - 12 = 33$ V.
Use $ΔEp=qΔV\Delta E_p = q\Delta V$ : $ΔEp=(3.0×10−6)(33)=9.9×10−5\Delta E_p = (3.0 \times 10^{-6})(33) = 9.9 \times 10^{-5}$ J. The positive sign means the ion’s electric potential energy increases.
The work done by the field is $Wfield=−ΔEp=−9.9×10−5W_{\text{field}} = -\Delta E_p = -9.9 \times 10^{-5}$ J. The electric field does negative work, so an external force would need to supply energy.

Common Mistake

Dropping the sign of the charge

Always include the sign of $q$ when using $ΔEp=qΔV\Delta E_p = q\Delta V$ . A negative charge, such as an electron, has the opposite potential energy change to a positive charge moving through the same potential difference.

Electrons, accelerating potential differences and electronvolts

A charged particle accelerated through a potential difference can gain kinetic energy. If it starts from rest and electrical energy is converted into kinetic energy, then:

ΔEk=qΔV\Delta E_k = q\Delta V

For many particle problems, you use the magnitude of the charge and the magnitude of the potential difference to find the energy gained.

The elementary charge, symbol $e$ , is the magnitude of the charge on a proton or electron:

\times 10^{-19}\ \text{C}

An electronvolt is a unit of energy. One electronvolt is the energy gained by an electron when it is accelerated through a potential difference of 1 V:

J1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}

Example

Energy gained by an electron

An electron is accelerated from rest through a potential difference of 2.0 kV. Find its kinetic energy in joules and in electronvolts.

Use the magnitude of the charge because the question asks for energy gained: $ΔEk=eΔV\Delta E_k = e\Delta V$ .
Substitute the values: $ΔEk=(1.60×10−19)(2.0×103)=3.2×10−16\Delta E_k = (1.60 \times 10^{-19})(2.0 \times 10^3) = 3.2 \times 10^{-16}$ J.
Since one electronvolt is $1.60 \times 10^{-19}$ J, the energy is 2.0 keV.

Uniform electric fields between parallel plates

A uniform electric field has the same field strength at every point. A good approximation is the field between two large, oppositely charged parallel plates, away from the edges.

The electric field points from the positive plate to the negative plate. The potential decreases in the direction of the electric field.

Parallel plates showing electric field lines and equipotentials

For parallel plates separated by distance $d$ with potential difference $ΔV\Delta V$ , the field strength magnitude is:

\frac{\Delta V}{d}

This shows why volts per metre, V m^-1, is equivalent to newtons per coulomb, N C^-1.

Definition

Equipotential

An equipotential is a line or surface where every point has the same electric potential. Moving a charge along an equipotential requires no work, because $ΔV=0\Delta V = 0$ .

Equipotential lines are always at right angles to electric field lines.

Example

Field strength between parallel plates

Two parallel plates have a potential difference of 600 V and are separated by 12 mm. Find the electric field strength and the force on a proton between the plates.

Convert the separation to metres and use $\Delta V/d$ : $\times 10^{-2}$ m, so $\times 10^{-2}) = 5.0 \times 10^4$ V m^-1.
Use $F = q E$ for a proton: $\times 10^{-19})(5.0 \times 10^4) = 8.0 \times 10^{-15}$ N.
The force is in the direction of the electric field, from the positive plate towards the negative plate.

Tip

Potential gradient

On a graph of potential against distance, the electric field strength is the negative gradient: $\approx -\Delta V/\Delta x$ . In many A-Level calculations you only need the magnitude, so use $\Delta V/d$ .

Common Mistake

Using E = V/d everywhere

The equation $\Delta V/d$ is for a uniform field, such as between parallel plates. Around a point charge, the field is not uniform, so you need the point-charge equations.

Potential around a point charge

A point charge is a charge treated as if it is concentrated at one point. This is a good model when the distance from the charge is much larger than the size of the charged object.

The electric potential at distance $r$ from a point charge $Q$ is:

\frac{Q}{4\pi\epsilon_0 r}

Here, $ϵ0\epsilon_0$ is the permittivity of free space. In practice, air is usually close enough to vacuum for A-Level calculations.

Point charge with radial field lines and concentric equipotentials

Important features:

Potential is positive around a positive charge.
Potential is negative around a negative charge.
Potential gets closer to zero as $r$ increases.
Potential follows a $1/ r$ relationship, whereas field strength follows a $1/r^2$ relationship.

The electric potential energy of a charge $q$ placed at a point where the potential is $V$ is:

E_p = qV

So for two point charges:

Ep=Qq4πϵ0rE_p = \frac{Qq}{4\pi\epsilon_0 r}

Example

Potential energy near a point charge

A charge of $+3.0 \times 10^{-6}$ C produces an electric potential. Find the potential 0.20 m from it, then find the electric potential energy of a charge of $−2.0 \times 10^{-6}$ C placed there.

Use $V = k Q / r$ , where $\times 10^9$ N m^2 C^-2: $\times 10^9)(3.0 \times 10^{-6})/0.20 = 1.35 \times 10^5$ V.
Use $E_p = qV$ : $Ep=(−2.0×10−6)(1.35×105)=−0.270E_p = (-2.0 \times 10^{-6})(1.35 \times 10^5) = -0.270$ J.
The negative value means the unlike charges have lower potential energy than when infinitely separated. You would need to supply 0.270 J to move the negative charge away to infinity slowly.

Adding potentials from more than one charge

Because electric potential is scalar, potentials from several charges add algebraically. That means you include signs, but you do not resolve components.

Vtotal=∑Q4πϵ0rV_{\text{total}} = \sum \frac{Q}{4\pi\epsilon_0 r}

This is much simpler than adding electric field strengths, which are vectors and require directions.

Example

Adding potentials from two charges

Two charges, $+4.0 \times 10^{-9}$ C and $−1.0 \times 10^{-9}$ C, are separated by 10 cm. Find the electric potential at the midpoint.

The midpoint is 0.050 m from each charge, so both contributions use the same distance but keep their signs.
Add the potentials algebraically: $Vtotal=(8.99×109)((4.0×10−9)/0.050+(−1.0×10−9)/0.050)=5.4×102V_{\text{total}} = (8.99 \times 10^9)((4.0 \times 10^{-9})/0.050 + (-1.0 \times 10^{-9})/0.050) = 5.4 \times 10^2$ V.
The answer is positive because the positive charge’s contribution is larger in magnitude than the negative charge’s contribution.

Linking potential, energy and motion

When a charged particle moves freely in an electric field:

A positive charge accelerates from higher potential to lower potential.
A negative charge accelerates from lower potential to higher potential.
Electric potential energy lost becomes kinetic energy gained, if no other forces do work.

This is just conservation of energy:

ΔEk=−ΔEp\Delta E_k = -\Delta E_p

So if $ΔEp\Delta E_p$ is negative, $ΔEk\Delta E_k$ is positive.

Exam technique

In the exam

Decide whether the field is uniform or radial: use $\Delta V/d$ for parallel plates, but $Q/(4\pi\epsilon_0 r)$ for point charges.
Keep signs for potential and potential energy calculations, especially when the moving charge is negative.
For multiple charges, add potentials as scalars; only use vector components when adding electric field strengths.

Self review

Check yourself

Why is electric potential a scalar, but electric field strength a vector?
A proton and an electron move through the same potential difference. How do their potential energy changes compare?
Why is no work done when a charge moves along an equipotential line?

Electric potential and energy

What you'll learn

The starting point: fields, work and charge

Electric potential

Potential is scalar

Potential difference and energy changes

Electrons, accelerating potential differences and electronvolts

Uniform electric fields between parallel plates

Potential around a point charge

Adding potentials from more than one charge

Linking potential, energy and motion

Electric potential and energy Revision Guide