Potential, potential energy and negative gradient
Key idea: H2 Physics lessons on point-charge interactions, uniform-field motion and capacitor energy.
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The core idea
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Big question: How do potential maps predict electric force and energy change?
Electric potential is potential energy per unit positive charge. A point charge gives V = Q/(4πε₀r), and a charge q has Ep = qV. Electric field is the negative potential gradient, so equipotentials are perpendicular to field lines and closer spacing indicates stronger field.
Build potential from work per charge
Electric potential at a point is the work done per unit charge by an external force in bringing a small positive test charge from infinity to that point. Equivalently, it is potential energy per unit positive charge. With zero at infinity, a point charge gives V = Q/(4πε₀r); the sign of Q remains. Potentials from several charges add algebraically.
A charge q at potential V has E_p = qV, and ΔE_p = qΔV. A negative charge can lose potential energy while moving to higher electric potential, so never infer energy change from ΔV without q.
Check your understanding: An electron moves through +200 V. What is the sign of its potential-energy change?
Negative: ΔE_p = qΔV and q is negative.
Read field from potential gradient
In one dimension, E = −dV/dx. The field points towards decreasing potential, and its magnitude equals the steepness of the V–x graph. A constant slope represents a uniform field.
Equipotential lines join equal V and are perpendicular to field lines. Moving along one requires no electric work. Close equipotential spacing shows a large potential change per distance and hence a strong field.
Check your understanding: A V–x graph slopes down steeply to the right. What is the field direction?
To the right, because E is the negative gradient; the negative slope gives positive E.
Key ideas to keep
- Potential is scalar and adds algebraically.
- A negative charge moves opposite to the electric field when only electric force acts.
- Moving along an equipotential changes neither V nor electric potential energy.
See the reasoning
Worked example
Keep potential, field and test-charge energy separate
Question: For a +4.0 nC source charge, compare V and E at 0.20 m and 0.40 m, then find Uᴱ for a −3.0 nC test charge at 0.20 m.
Step 1: Compare the two distance laws
Why: Potential is inverse-distance while field is inverse-square.
Working: Doubling r halves V but quarters E.
Step 2: Find the stated potential
Why: V is determined by the source charge alone.
Working: At 0.20 m from +4.0 nC, V = kQ/r = 180 V.
Step 3: Introduce the test charge
Why: Potential energy is qV and retains both signs.
Working: U = (−3.0×10⁻⁹)(180) = −5.39×10⁻⁷ J.
Answer: V = kQ/r, so doubling r halves V: 180 V to 89.9 V. E = kQ/r², so it quarters: 899 N C⁻¹ to 225 N C⁻¹. Uᴱ = qV = (−3.0 × 10⁻⁹)(180) = −5.39 × 10⁻⁷ J.
Check: A negative potential energy is expected for unlike source and test charges with zero at infinity.
Use a hint if needed
Practise with support
Try this
Potential falls linearly from 120 V to 40 V over +0.20 m. Find the signed field component.
Hint: Calculate the signed potential gradient before applying the minus sign.
Check your answer
dV/dx = (40 − 120)/0.20 = −400 V m⁻¹, so Eₓ = −dV/dx = +400 V m⁻¹.
Now work without the hint
Practise independently
Your turn
Define electric potential using external work, then relate V, Uᴱ and E for a point charge.
Check your answer
V is external work per unit charge in bringing a small positive test charge slowly from infinity to the point. For source Q, V = Q/(4πε₀r); a two-charge system has Uᴱ = qV = Qq/(4πε₀r). Locally, E = −dV/dr, so the field points down the potential gradient.
Avoid these traps
Common mistakes
Common mistake
Potential contributions must be added as vectors.
What is wrong with this reasoning?
Show better thinking
Potential is scalar, so contributions add algebraically; field contributions add as vectors.
Common mistake
Electric field points towards increasing potential.
What is wrong with this reasoning?
Show better thinking
E is the negative potential gradient and points towards decreasing potential.
Common mistake
Work done by the field is qΔV.
What is wrong with this reasoning?
Show better thinking
ΔU = qΔV, while work done by the electric field is −ΔU.
Write for the examiner
Exam guidance
Track both signs in ΔEp = qΔV before deciding whether kinetic energy rises.
Exam-style practice [8 marks]
Define V, then find V and Uᴱ at 0.25 m from +6.0 nC for a −2.0 nC test charge. State the field direction from E = −dV/dr.
Plan before you answer
- Define V in words.
- Calculate source potential, then qV.
- Use the negative gradient for field direction.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
V is external work per unit positive charge from infinity. V = 216 V and Uᴱ = qV = −4.32 × 10⁻⁷ J. V decreases with increasing r, so −dV/dr is positive radial: the field points outward.
Come back in three days
Check what stayed with you
Recall question
A +3.0 nC charge moves through ΔV = −50 V. Find ΔU and work done by the field.
Check the answer
ΔU = qΔV = −1.50 × 10⁻⁷ J. Work done by the field is −ΔU = +1.50 × 10⁻⁷ J.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 14 states no explicit exclusions. Coulomb-force, point-field and point-potential equations are used for point charges in free space or air. Electric field and force are vectors; potential and potential energy are scalars. E = ΔV/d is restricted to a uniform field, while E = −dV/dr is the local negative potential gradient.
- GCE A-Level H2 PhysicsTopic 14(c) / Topic 14(d) / Topic 14(e) / Topic 14(f) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027