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Electric potential
Electric potential carries over from gravitational potential almost word for word: zero at infinity, work as charge times potential difference, and equipotentials along which no work is done. Two things differ. The sign depends on the sign of the charge, and the data booklet's notation needs handling carefully.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
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, gradient in one direction and area in the other.
COMMON MISCONCEPTION
A volt is an amount of energy.
Joules 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; around a negative charge it is negative. Gravity, with one sign of mass only, never had that choice.
A volt is a joule per coulomb, so potential and potential difference measure energy per unit charge, and no energy changes hands until a charge actually moves. The energy transferred in moving a charge Q through a potential difference is
which is also the equation sitting quietly underneath the electronvolt you met in the quantum unit, the energy of one electron moved through one volt.
Equipotentials, again
Surfaces of constant potential ring a point charge exactly as they ring a planet, they cross the field lines at right angles, and no work is done moving a charge along an equipotential. The picture transfers from the gravitational lesson with a single edit. For a positive central charge the field arrows point outward, and moving inward makes the potential more positive instead of more negative.
OCR adds one tidy corollary. Outside itself a charged conducting sphere behaves as a point charge at its centre, so a sphere of radius R holding charge Q sits at potential V = Q/4πε0R, and the charge it holds is proportional to that potential. The ratio Q/V is its capacitance, C = 4πε0R, so the larger the sphere the more charge each volt stores.
The graphs and the dictionary
Plot both quantities against distance for a positive charge and the two curves make the family resemblance exact: V falls as 1/r, E falls as 1/r2, so the field falls away faster than the potential. The translation rules between them are the ones you already own:
with the size of the gradient of the V graph giving the field strength, and the area under the E against r graph between two distances giving ΔV. Read that equation as a statement about magnitudes, because the signs do not sit still. Around the positive charge above, V falls as r grows, so ΔV/Δr is negative while E points outwards and is counted positive. Written with its sign the relation is E = −ΔV/Δr, the field pointing down the potential gradient, from high potential towards low.
The booklet prints the electric line without that minus sign, quoting E as a magnitude, where the gravitational version g = −ΔV/Δr keeps it. No physics differs between the two, only the printed convention. So take the steepness of the V against r graph for the size of E, work the direction out from the field itself, and follow the booklet's form for whichever field you are working in.
WORKED EXAMPLE
Work along and across the map
A +2.0 μC charge moves from a point at a potential of 400 V to a point at 150 V. Find the work done on it by the field, then say what changes if it moves the other way, and find the work needed to move it along an equipotential at 400 V.
Fix the direction before touching the arithmetic. ΔV means the rise in potential, final minus initial, so ΔV = 150 − 400 = −250 V. The work an outside agent must supply is QΔV, and the work done by the field is the negative of it, Wfield = −QΔV.
So Wfield = −(2.0 × 10−6) × (−250) = +5.0 × 10−4 J. Positive, and it has to be: a positive charge running downhill from 400 V to 150 V is pushed the way it is already going, so the field delivers energy to it.
Send it back, from 150 V to 400 V, and every sign turns over. ΔV = +250 V, the field does −5.0 × 10−4 J and an outside agent has to supply 5.0 × 10−4 J to push the charge uphill. The size of the answer never depended on the direction; the sign did, and only the words “from” and “to” set it.
Along an equipotential ΔV = 0, so the work is zero, however long the path.
The energy transferred depends only on the contour you start and finish on, never the route between them.
GUIDED PRACTICE
Halving by walking away
The potential at distance r from a point charge is 60 V. Without computing the charge, find the potential at 2r and at 4r.
Show the working
V falls as 1/r, not as 1/r2, so at 2r it reads 30 V and at 4r 15 V.
The inverse square belongs to the field strength. Potential keeps the gentler slope, so its graph creeps down towards zero far more reluctantly than the field does.
ASSESSMENT FOCUS
- Define potential with all three ingredients. Work done, per unit positive charge, brought from infinity. The electric version needs that sign spelt out; the gravitational one never does.
- ΔW = QΔV runs both ways. Work is supplied when a positive charge climbs to a higher potential and delivered when it falls, so track the signs of Q and ΔV together.
- No work along an equipotential, and field lines crossing equipotentials at right angles. Two stock marks, on offer whenever the diagram appears.
- For the graph translation, the size of the gradient of V against r gives E and the area under E against r gives ΔV. Quote whichever direction the data supports.
- Follow the booklet's signs exactly. Gravity keeps its minus in g = −ΔV/Δr, while the electric E = ΔV/Δr is printed as a magnitude. Say "magnitude of the gradient" when you quote it, since the potential round a positive charge falls as r grows and the raw gradient is negative. Mixing the two conventions costs a mark almost every year.
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 joules per coulomb times the coulombs.
Show the answer
= (8.99 × 109 × 2.0 × 10−9) / 0.30 = 60 V.
Coming from infinity, ΔV = 60 V, so = 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 the whole way.
Potential is work per coulomb from infinity, and its sign follows the charge.
The size of the V against r gradient gives E, and the field points down the gradient; the area under E against r gives ΔV.
A volt is a joule per coulomb, never an amount of energy on its own.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the electric potential questions page.
WHERE TO GO NEXT
- Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- 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, gradient in one direction and area in the other.
Open the full revision checklist to track your progress across the whole unit.