AQA A-Level Physics Paper 2, June 2025: Question 16

📋 What This Question Tests

This multiple-choice question assesses your fundamental understanding of electric fields and potential gradients:

The nature and uniformity of the electric field between two oppositely charged or differing potential parallel plates.
The mathematical definition of electric field strength as potential gradient: E = −ΔV / Δx .
Interpreting field direction relative to a defined spatial coordinate axis (vector sign conventions).
Distinguishing between graphs of electric potential ( V ) versus distance and electric field strength ( E ) versus distance.

Option D

Graph D shows a horizontal line below the zero axis (a constant negative value).

Mark Scheme: 1 mark for selecting D (AO1 - recall and understanding of uniform field characteristics).

Uniform Field: The electric field between two large, parallel conducting plates is uniform (constant magnitude and constant direction everywhere between the plates, excluding edge effects).
Field Direction: Electric field lines always point from higher potential to lower potential.
Potential Gradient: Electric field strength is defined as the negative gradient of electric potential with respect to distance:
E = −(dV / dx)

Determine whether the field is constant:
Because the plates are parallel and conducting, the field is uniform. Therefore, the magnitude of E does not depend on position x . This requires E to be a horizontal line (constant value).
→ Immediately eliminates graphs A, B, and C without doing any calculation!
Identify potentials at both boundaries:
Bottom plate (at x = 0 ): V = −20 V
Top plate: V = −10 V
Note: −10 V > −20 V , so the top plate is at a higher potential than the bottom plate.
Determine the direction and sign of E:
Electric field lines point from higher potential to lower potential: from the top plate (−10 V) down to the bottom plate (−20 V).
The displacement coordinate x is defined as pointing upwards.
Since the electric field vector points downwards (opposite to the positive x direction), E must be negative.
Mathematical check via potential gradient:
ΔV = V_top − V_bottom = (−10) − (−20) = +10 V
Since Δx > 0 (upwards), the gradient ΔV / Δx is positive.
Therefore: E = −(ΔV / Δx) = −(positive value) = negative constant .
This matches Graph D precisely.

Elimination First: As soon as you see parallel plates, ask yourself: "Is the field uniform or radial?" Parallel plates = uniform field = horizontal graph for E vs x . You could choose D in under 10 seconds!
Watch the Signs of Negative Numbers: Remember that −10 is greater than −20 . Treat potentials algebraically, not just by magnitude.
Read the Axes Carefully: Confirm whether the vertical axis is V (electric potential) or E (electric field strength). If it were V vs x , the line would be straight with a positive slope!

Confusing E with V: Choosing graph A because students know potential increases linearly from −20 V to −10 V . A graph of V against x would look like A, but the question asks for E !
Assuming Radial Field Variation: Choosing B or C because students wrongly apply the inverse-square law ( E ∝ 1/r² ) from point charges, forgetting parallel plates produce a uniform field.
Negative Number Trap: Thinking −20 V is "larger" than −10 V , thus predicting the field points upwards and expecting a positive E value.

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