The big idea: Electric potential V tells you the potential energy per unit positive charge at a point in a field.
Two things to know
- It is a single number (a scalar) — no direction — so you just add potentials, never worry about components.
- For a point charge Q it falls off as 1/r: double the distance and the potential halves.
- electric potential (V, i.e. J C⁻¹)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻²
- charge creating the field (C)
- distance from the charge (m)
Potential ≠ field: The field E is a vector (a push, in N C⁻¹) and falls off as 1/r². The potential V is a scalar (an energy-per-charge, in J C⁻¹ = volts) and falls off as 1/r. Don't mix the two up.
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This is the big HL difference from gravity. Mass only attracts, so gravitational potential is always negative. But charge comes in two signs, so electric potential can be positive or negative — and the sign carries real meaning.
Near a POSITIVE charge
- Q > 0, so V = kQ/r is positive
- V is highest close to the charge and falls toward 0 far away
- A positive test charge has positive PE — it is 'uphill'
- Release a positive charge ⇒ it rolls to lower V (away)
Near a NEGATIVE charge
- Q < 0, so V = kQ/r is negative
- V is most negative close to the charge and rises toward 0 far away
- A positive test charge sits in a 'potential well'
- Release a positive charge ⇒ it falls to lower (more negative) V (toward it)
The convention: Potential is defined to be zero at infinity. So V at a point is the work done per coulomb to bring a small positive charge from infinity to that point. Positive V means you had to push it in; negative V means it was pulled in.
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From potential to energy: Put a second charge q at a point where the potential is V and it has electric potential energy Ep = qV. For two point charges this becomes Ep = kQq/r — a single scalar for the pair.
- electric potential energy of the pair (J)
- the two charges (C), with sign
- separation of the charges (m)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻²
Sign of the energy: Like charges (Qq > 0) give positive Ep — energy was stored pushing them together; they fly apart if released. Unlike charges (Qq < 0) give negative Ep — a bound system, like an electron near a proton.
Worked example — potential from a point charge
Find the electric potential at a point 0.30 m from a small charge Q = 2.0 × 10⁻⁹ C.
Solution
- Write the given formula first:
- Substitute the values (k = 8.99 × 10⁹):
- Work it out — keep the unit:
Final answer
V = 60 V (positive, because Q is positive).
Only the endpoints matter: To move a charge q between two points you do work equal to its change in PE.
Why the route does not matter
- PE depends only on where you are, not the route.
- So the work is path-independent.
- It depends only on the potential difference ΔV between start and finish.
- work done on the charge (J)
- the charge being moved (C), with sign
- potential difference, V_final − V_initial (V)
Worked example — work to move a charge
A charge q = 1.5 × 10⁻⁹ C sits at the point above where V = 60 V. It is moved to a point where V = 20 V. Find the work done.
Solution
- Write the given formula first:
- Find ΔV = Vfinal − Vinitial:
- Substitute and work it out — keep the unit:
Final answer
W = −6.0 × 10⁻⁸ J. The negative sign means the field did the work — energy is released, not supplied.
Mind the sign: Always take ΔV = Vfinal − Vinitial. A negative W means the field pushed the charge along (energy released); a positive W means you had to push against the field (energy supplied).
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The field is the slope of the potential: Move a small distance Δr along a field line and the potential changes by ΔV. The field strength is how steeply V changes.
- electric field strength (V m⁻¹ = N C⁻¹)
- change in electric potential (V)
- distance moved (m)
What the minus sign says: The field points the way the potential falls: from high V to low V.
On a V–r graph, the field is minus the gradient. Between parallel plates the graph is a straight line, so E = V/d.
Equipotentials
An equipotential is a line (or surface) joining points at the same potential. Drawn at equal steps of V, they show the field as clearly as field lines do.
Worked example — field from a potential graph
Along a line between two plates the potential falls steadily from 300 V to 100 V over 0.040 m. Find the electric field strength.
Solution
- Use the gradient:
- Work it out:
Final answer
E = 5.0 × 10³ V m⁻¹, pointing the way V falls (towards the 100 V end).
Where it shows up: Electric potential and work are HL only (D.2). They come up in Paper 1A and Paper 2.
| Paper | What they ask |
|---|---|
| Paper 1A | A one-step 'what is V?', 'is V positive or negative here?', or 'compare V at two distances' |
| Paper 2 | Determine the work to move a charge using W = qΔV, often after first computing V at two points with V = kQ/r and adding scalars |
Three easy marks: (1) V is a scalar — just add the potentials from several charges (with sign). (2) Carry the sign of every charge into kQ/r. (3) Work uses ΔV = Vfinal − Vinitial, so a charge moving to lower V for the same sign gives negative work.
IB-style question — potential between two charges
Charges of +4.0 × 10⁻⁹ C and −4.0 × 10⁻⁹ C are 0.40 m apart. Determine the electric potential at the midpoint between them.
Solution
- Potential is a scalar: Vtotal = V₁ + V₂, each from V = kQ/r. The midpoint is 0.20 m from each charge.
- Substitute (the two charges are equal and opposite, both at r = 0.20 m):
- The two terms cancel exactly:
Final answer
V = 0 V at the midpoint — equal and opposite charges give zero potential there (even though the field is not zero).