Physics 9702 · Paper 4

Electric Potential

The “voltage” you have used since AS finally gets its full definition — the same work-from-infinity idea as gravitational potential, but with signs that matter.

The definition (two marks)

The at a point is the work done per unit charge in moving a positive charge from infinity to the point. Split exactly there in three sessions (9702/41/O/N/23 Q5(a)) (9702/42/M/J/24 Q5(a)) (9702/41/O/N/25 Q5(a)) — keep “positive” and “from infinity”.

V = \dfrac{Q}{4\pi\varepsilon_0 r} \qquad E_P = \dfrac{Qq}{4\pi\varepsilon_0 r}

Symbols

$V$ = potential at distance r from charge Q (V)
$E_P$ = potential energy of two point charges Q and q (J)
$r$ = distance from the charge (single power!) (m)

Both formulas divide by $r$ , not $r^2$ — a one-marker has hung on exactly this (9702/42/F/M/23 Q4(b)(iii)). Unlike gravitational potential (always negative), $V$ takes the sign of the charge: positive charges make potential hills, negative charges make wells.
Potential is a scalar. With several charges, add the values with their signs — no directions, no components. That is why examiners love mixing potential and field questions: one adds like numbers, the other like arrows.

Potential along a proton–electron line (the 2025 exam curve, sampled from V ∝ 1/x − 1/(120 − x)): +130 V at 10 pm, +32 V at 30 pm, zero exactly at the 60 pm midpoint — signed scalars adding, no vectors needed.

Worked example

Along a hydrogen atom (2025 paper)

A proton and an electron sit 120 pm apart. Find the total potential 30 pm from the proton (9702/41/O/N/25 Q5(b)(ii)).

Two signed contributions: $V = \dfrac{e}{4\pi\varepsilon_0}\left(\dfrac{1}{30\ \text{pm}} - \dfrac{1}{90\ \text{pm}}\right)$ .
$V = 1.44\times10^{-9} \times \left(\dfrac{1}{30} - \dfrac{1}{90}\right)\times10^{12} = +32\ \text{V}$ .

Answer

V \approx +32\ \text{V}

Because the charges have equal size, $V = 0$ exactly at the midpoint — the figure's zero crossing at 60 pm.

Capacitance preview

A 2025 part chained an isolated sphere's capacitance into this lesson's formula: from

C = Q/V

, a sphere holding 83 pC at 1.2 V, then

r = Q/4\pi\varepsilon_0 V = 0.62

m (9702/42/O/N/25 Q6(b)). The capacitance chapter builds on exactly this

V = Q/4\pi\varepsilon_0 r

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