Physics 9702 · Paper 4

The Field of a Point Charge

Divide Coulomb's force by the test charge and the field itself appears — an inverse-square law with a vector personality.

E = Q/4πε₀r²

E = \dfrac{Q}{4\pi\varepsilon_0 r^2}

Symbols

$E$ = field strength at distance r from the charge (N C⁻¹ (= V m⁻¹))
$Q$ = the charge producing the field (C)
$r$ = distance from the charge (or sphere centre) (m)

It is Coulomb's law with one charge removed: the force $Q_1Q_2/4\pi\varepsilon_0 r^2$ per unit of $Q_2$ . Double the distance, quarter the field — the same inverse square as gravity.

E–x for a charged conducting sphere: exactly zero everywhere inside, maximum at the surface, then the sampled inverse square E ∝ 1/x² outside. Exam graphs are read backwards: the radius is where E stops being zero.

Exam graphs run this backwards: given $E$ against distance for a charged sphere, the radius is where the field stops being zero, and one point on the curve gives the charge — $Q = E \cdot 4\pi\varepsilon_0 x^2 = 2.5\times10^{-8}$ C from the 2024 graph (9702/42/O/N/24 Q6(c)).
Fields are vectors. With two charges, work out each field's magnitude and direction at the point, then combine. A 2024 question placed P between opposite charges: both fields pointed the same way, so the magnitudes added to $3Q/8\pi\varepsilon_0 x^2$ (9702/41/O/N/24 Q5(c)(iii)) — students who subtracted saw “opposite charges” and guessed. Draw the two arrows first, every time.

Your turn— tap to reveal the worked answer (9702/42/O/N/25 Q6(b))

An isolated conducting sphere stores +83 pC and sits at a potential of 1.2 V. Its radius works out (next lesson) as 0.62 m. What is the field strength at its surface?

Treat the charge as a point at the centre: $E = \dfrac{83\times10^{-12}}{4\pi\varepsilon_0 \times 0.62^2} = 1.9\ \text{N C}^{-1}$ — and the unit itself carries a mark.

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