AQA A-Level Physics Paper 2, June 2023: Question 17
1 mark · Medium difficulty · Multiple Choice
Identify what the area under the curve of an electric field strength versus distance graph from the surface of a charged sphere to infinity represents.
Practise this questionQuestion
Question text
17 The graph shows the variation of electric field strength E surrounding a charged
sphere of radius R. The distance from the centre of the sphere is r.
The total area under the curve from R to infinity is
[1 mark]
A the capacitance of the sphere.
B the charge held on the sphere.
C the electric potential of the sphere.
D the energy needed to remove an electron from the sphere.
Mark scheme
Show the mark scheme
17 C the electric potential of the sphere.
How to answer it
Area Under an Electric Field Strength–Distance Graph
This question assesses your understanding of graphical relationships in non-uniform radial electric fields, specifically:
- The relationship between electric field strength ( E ) and electric potential ( V ).
- Physical interpretation of the area under an E – r curve ( E = -ΔV / Δr ).
- The definition of electric potential at the surface of a charged sphere relative to infinity where V = 0 .
Identifying the Physical Quantity Represented by the Area
Relating Field Strength E to Potential V
✅ Correct Answer
C: the electric potential of the sphere.
🧠 Exam Technique: Graphical Calculus
Always inspect the gradient and area using calculus definitions:
- Gradient of a y vs x graph = dy / dx
- Area under a y vs x graph = ∫ y dx
Here, the area is ∫ E dr . Check the units: (N C⁻¹) × m = N m C⁻¹ = J C⁻¹ = V (volts).
📐 Step-by-Step Mathematical Derivation
- Recall the field-potential relationship:
The electric field strength is equal to the negative potential gradient:
E = - dV / dr or ΔV = - ∫ E dr - Integrate from the surface (r = R) to infinity (r = ∞):
Area = ∫R∞ E dr - Substitute the field of a point charge / conducting sphere:
E = Q / (4πε₀r²)
Area = ∫R∞ [Q / (4πε₀r²)] dr = [-Q / (4πε₀r)]R∞ = 0 - (-Q / (4πε₀R)) = Q / (4πε₀R) - Compare with standard definitions:
The electric potential V on the surface of a sphere of radius R is precisely:
V = Q / (4πε₀R)
Since V = 0 at infinity, the area represents the magnitude of the electric potential of the sphere.
💡 Key Knowledge: Radial Fields
- Inside a charged conducting sphere: E = 0 , while V is constant and equal to its surface value.
- At surface & beyond (r ≥ R): The sphere acts as if all its charge Q is concentrated at its centre.
- Potential Definition: Work done per unit positive charge in bringing a small test charge from infinity to that point. Hence:
V = - ∫∞r E dr = ∫r∞ E dr
❌ Common Distractors & Why They Are Wrong
- A (Capacitance): Capacitance is C = Q / V . For an isolated sphere, C = 4πε₀R , which is a constant geometry property, not the area under an E–r graph.
- B (Charge held on sphere): Charge Q has units of coulombs (C). The area has units of (V m⁻¹) × m = V (volts or J C⁻¹), so dimensionally it cannot be charge.
- D (Energy needed to remove an electron): Energy requires multiplying electric potential by charge ( ΔEₚ = qΔV ). The area is work done per unit charge, not total energy for an electron.
Topics
Physics · 3.7 Fields and their consequences (A-level only)
Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.