Electric potential (A-level only)

AQA·Physics·3.7.3.3

Electric Potential and Potential Difference

A charge has electric potential energy because of its position in an electric field. Electric potential describes the electric potential energy per unit charge at a point. If a small test charge $$q$$ has electric potential energy $$U$$ at that point, the potential is:

V=UqV = \frac{U}{q}

The value of a potential depends on the reference chosen for zero. Absolute electric potential uses zero potential at infinity. It can be defined as the work done per unit positive charge by an external agent bringing a small positive test charge from infinity to the point, without changing its kinetic energy. The test charge is taken to be small so that it does not significantly alter the field being measured.

Potential difference and energy change

The potential difference between two points is the change in electric potential energy per unit charge when a charge moves between them. It is the difference between the potential at the final point and the potential at the initial point.

For a charge $$Q$$ moving through a potential difference $$\Delta V$$, the change in its electric potential energy, $$\Delta W$$, is:

ΔW=QΔV\Delta W = Q\Delta V

The signs matter: a positive charge gains potential energy when it moves to a higher potential, while a negative charge loses potential energy for the same change in potential.

If an external agent moves the charge without changing its kinetic energy, the work done by that agent equals the change in potential energy, $$\Delta W$$. The work done by the electric field is the negative of that change, $$-\Delta W$$.

Both potential and potential difference are measured in volts. One volt is one joule per coulomb: $$1,\mathrm{V} = 1,\mathrm{J,C^{-1}}$$.

Equipotential Surfaces

An equipotential surface is a surface on which electric potential has the same value at every point. It need not be a physical surface: it represents a set of points with the same potential.

If a charge moves between two points on the same equipotential surface, the change in potential is $$\Delta V = 0$$. Its change in electric potential energy is therefore:

ΔU=QΔV=0\Delta U = Q\Delta V = 0

Here, $$Q$$ is the charge and $$\Delta U$$ is its change in electric potential energy. The electric field does no work on a charge moving along the surface.

Equipotential surfaces and electric field lines

Electric field lines are perpendicular to equipotential surfaces. If the field had a component along a surface, it would do work on a charge moving a small distance in that direction. The potential would then change along the surface, contradicting the fact that the surface is equipotential.

Electric Potential and Field in a Radial Field

A point charge produces a radial electric field. At a distance $$r$$ from the charge, the electric potential describes the electric potential energy per unit charge. The potential is defined relative to a reference level: here, it is taken to be zero at infinity.

Potential around a point charge

For a point charge $$Q$$, the potential at distance $$r$$ is:

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

Here, $$\varepsilon_0$$ is the permittivity of free space. The potential has the same sign as the source charge: it is positive around a positive charge and negative around a negative charge. Its magnitude, regardless of sign, is:

∣V∣=14πε0∣Q∣r|V| = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r}

As distance increases, the magnitude of the potential decreases towards zero. On a graph of signed $$V$$ against $$r$$, the curve lies above the horizontal axis for a positive source charge and below it for a negative source charge. Both curves approach zero as $$r$$ increases.

Potential and field strength with distance

The magnitude of the electric field strength at distance $$r$$ from the same point charge is:

E=14πε0∣Q∣r2E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r^2}

Potential magnitude follows an inverse-distance relationship: $$|V| \propto 1/r$$. Field strength follows an inverse-square relationship: $$E \propto 1/r^2$$. Graphs of both magnitudes against distance decrease towards zero, but field strength decreases more rapidly with distance.

For example, doubling the distance makes $$|V|$$ half as large, while $$E$$ becomes one-quarter as large.

Electric Field as a Potential Gradient

Electric field strength depends on how quickly electric potential changes with position. A steep change in potential over a short distance corresponds to a strong field; a gradual change corresponds to a weaker field.

Along a radial coordinate $$r$$, the signed radial field component $$E_r$$ is related to the potential $$V$$ by:

Er=−dVdrE_r = -\frac{dV}{dr}

The derivative $$dV/dr$$ is the potential gradient: the rate at which potential changes as $$r$$ increases. The minus sign shows that the field points towards decreasing potential. If potential decreases as $$r$$ increases, $$E_r$$ is positive; if potential increases as $$r$$ increases, $$E_r$$ is negative.

Where the field and displacement lie along the same line, the field magnitude at a point is $$|dV/dr|$$. Over an interval, the magnitude of the average potential gradient is $$|\Delta V/\Delta r|$$. This gives the average field magnitude over the interval when the gradient retains the same sign.

Potential difference from a field graph

The potential change from radial position $$r_1$$ to $$r_2$$ can be found by adding the contributions from the field along the path:

ΔV=V(r2)−V(r1)=−∫r1r2Er dr\Delta V = V(r_2)-V(r_1) = -\int_{r_1}^{r_2} E_r\,dr

On a graph of $$E_r$$ against $$r$$, the integral is the signed area between the graph and the radial axis, taken in the direction from $$r_1$$ to $$r_2$$. The potential change is the negative of that signed area. Areas above and below the axis have opposite signs and must be combined accordingly.

If a graph shows field magnitude rather than the signed component, and the field has a constant direction throughout the region, the area gives the magnitude of the potential difference. Moving along the field gives a decrease in potential; moving against it gives an increase.

For example, moving $$3,\mathrm{m}$$ along a constant field of magnitude $$2,\mathrm{V,m^{-1}}$$ gives a potential change of $$-6,\mathrm{V}$$. Travelling back over the same interval gives a change of $$+6,\mathrm{V}$$.