Physics › Electric fields › Electric potential
Electric potential
The potential idea crosses over from gravity almost word for word: zero at infinity, work as charge times potential difference, equipotentials where movement is free. The differences are the sign, which now depends on the charge, and one quirk of notation AQA wants handled exactly.
Builds on Coulomb's law and electric field strength and Gravitational potential.
IN THIS TOPIC
- Define absolute electric potential with its zero at infinity, and use ΔW = QΔV.
- Use the radial potential of a point charge, and read equipotential diagrams.
- Translate between the E and V graphs: E = ΔV/Δr, and ΔV as the area under E against r.
WHAT YOU PROBABLY THINK
A volt is an amount of energy.
Potential, priced per coulomb
The absolute electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point, with the zero at infinity exactly as for gravity. In the radial field of a point charge,
and here the sign takes care of itself: around a positive charge the potential is positive, because pushing another positive charge inward costs work, while around a negative charge it is negative. Gravity, with only one sign of mass, never had the choice.
The lie above is worth killing precisely. A volt is a joule per coulomb: potential and potential difference price energy per unit charge, and no energy changes hands until a charge actually moves. The energy bill for moving a charge Q through a potential difference is
which is also the equation quietly underneath the electronvolt from the quantum unit: one electron moved through one volt.
Equipotentials, again
Surfaces of constant potential ring a point charge exactly as they ring a planet, crossing the field lines at right angles, and no work is done moving a charge along an equipotential. The whole picture transfers from the gravitational lesson with a single edit: for a positive central charge the field arrows point outward, and the potential grows more positive as you move in rather than more negative.
The graphs and the dictionary
Plot both quantities against distance for a positive charge and the two curves tell the family story: V falls as 1/r, E falls as 1/r2, so the field dies away faster than the potential. The translation rules between them are the ones you already own:
with the gradient of the V graph giving the field strength, and the area under the E against r graph between two distances giving ΔV. One notational point deserves care: AQA prints the electric relation without a minus sign, quoting E as a magnitude, where the gravitational version g = −ΔV/Δr keeps its sign. The physics has not changed, since the field still points from high potential toward low for a positive test charge; only the printed convention differs, and your answers should follow the booklet's form for each field.
THE EXAM BIT
- Define potential with all three ingredients: work done, per unit positive charge, from infinity. The electric definition needs the charge's sign specified.
- ΔW = QΔV runs both ways: supplied work when a positive charge climbs to higher potential, delivered energy when it falls. Track the signs of both Q and ΔV.
- No work along an equipotential, and field lines cross equipotentials at right angles: two stock marks whenever the diagram appears.
- Graph translation: gradient of V against r gives E; area under E against r gives ΔV. Quote the direction the data supports.
- Follow the booklet's signs exactly: g = −ΔV/Δr for gravity, E = ΔV/Δr as a magnitude for electric fields. Mixing the conventions is a classic dropped mark.
CHECK YOURSELF
A point charge of +2.0 nC sits in air. Find the potential 0.30 m away, and the work needed to bring a +1.5 nC charge from far away to that point.
Show a hint
Radial potential first; then the price per coulomb times the coulombs.
Show the answer
V = Q/4πε0r = (8.99 × 109 × 2.0 × 10−9) / 0.30 = 60 V.
Coming from infinity, ΔV = 60 V, so ΔW = QΔV = 1.5 × 10−9 × 60 = 9.0 × 10−8 J.
Positive work, supplied by whatever does the pushing: both charges are positive, so the arrival is uphill all the way.
Potential is work per coulomb from infinity; the sign follows the charge.
Gradient gives E from V; area gives ΔV from E.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.