Physics 9702 · Paper 4

Field = −Potential Gradient

Field and potential are two descriptions of one landscape: potential is the height map, field is the slope. The minus sign in between is worth its own mark.

E = −dV/dx

E = -\dfrac{\Delta V}{\Delta x}

Symbols

$E$ = field strength at the point (V m⁻¹)
$\Delta V / \Delta x$ = gradient of the potential–distance graph (V m⁻¹)

Stated as words: field strength equals the potential gradient (M1), with a minus sign (A1) — the 2024 mark scheme splits it exactly so (9702/41/O/N/24 Q5(a)). The minus says the field pushes positive charges downhill in potential.
On any $V$ – $x$ graph, the field at a point is minus the tangent's gradient. Steep potential = strong field; flat potential = zero field. That instantly explains conductors: $V$ is constant inside, so $E = 0$ inside.

V–x between two charged conducting spheres. The exam skill is reading it backwards: flat sections give each sphere's radius (V is constant inside a conductor), the signs of V near each surface give the charge signs.

The recycled “three conclusions” question hands you exactly this graph for two charged spheres and marks any three of: the radii (where the curve is flat), the charge signs (sign of $V$ near each surface), and whether the magnitudes are equal (symmetry of the curve) (9702/41/O/N/23 Q5(b)) (9702/42/M/J/24 Q5(b)).

Zero potential vs zero field

Between two charges, $V = 0$ and $E = 0$ are different demands: $V = 0$ needs opposite signs (a positive and a negative potential cancelling), while $E = 0$ between the charges needs the same sign (two fields pointing opposite ways). The charges cannot be both — so no point between them can have both zero at once. That argument, in sentences, was a full three-marker (9702/41/O/N/24 Q5(b)).
What a released charge does at each: at an $E = 0$ point (between like charges) it feels no force and stays at rest — writing “it moves away” lost both marks in 2024 (9702/42/M/J/24 Q5(c)). Released off-balance between opposite charges, it accelerates towards the attractor with increasing acceleration as the inverse-square force grows (9702/41/O/N/23 Q5(c)).

Field is minus the slope of potential. Flat $V$ ⇒ zero $E$ . Between charges: $V=0$ needs opposite signs, $E=0$ needs like signs — never both.

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Field = −Potential Gradient

Field and potential are two descriptions of one landscape: potential is the height map, field is the slope. The minus sign in between is worth its own mark.

E = −dV/dx

E = -\dfrac{\Delta V}{\Delta x}

Symbols

$E$ = field strength at the point (V m⁻¹)
$\Delta V / \Delta x$ = gradient of the potential–distance graph (V m⁻¹)

Stated as words: field strength equals the potential gradient (M1), with a minus sign (A1) — the 2024 mark scheme splits it exactly so (9702/41/O/N/24 Q5(a)). The minus says the field pushes positive charges downhill in potential.
On any $V$ – $x$ graph, the field at a point is minus the tangent's gradient. Steep potential = strong field; flat potential = zero field. That instantly explains conductors: $V$ is constant inside, so $E = 0$ inside.

The recycled “three conclusions” question hands you exactly this graph for two charged spheres and marks any three of: the radii (where the curve is flat), the charge signs (sign of $V$ near each surface), and whether the magnitudes are equal (symmetry of the curve) (9702/41/O/N/23 Q5(b)) (9702/42/M/J/24 Q5(b)).

Zero potential vs zero field

Between two charges, $V = 0$ and $E = 0$ are different demands: $V = 0$ needs opposite signs (a positive and a negative potential cancelling), while $E = 0$ between the charges needs the same sign (two fields pointing opposite ways). The charges cannot be both — so no point between them can have both zero at once. That argument, in sentences, was a full three-marker (9702/41/O/N/24 Q5(b)).
What a released charge does at each: at an $E = 0$ point (between like charges) it feels no force and stays at rest — writing “it moves away” lost both marks in 2024 (9702/42/M/J/24 Q5(c)). Released off-balance between opposite charges, it accelerates towards the attractor with increasing acceleration as the inverse-square force grows (9702/41/O/N/23 Q5(c)).

Field is minus the slope of potential. Flat $V$ ⇒ zero $E$ . Between charges: $V=0$ needs opposite signs, $E=0$ needs like signs — never both.