Suppose points A and B are separated by displacement Δr. If their electric potentials are VA and VB, the potential difference is the final potential minus the initial potential.
Key equation
Δ V=VB-VA
V_A
potential at the initial point A (V)
V_B
potential at the final point B (V)
Potential gradient is the change in electric potential per unit displacement along a chosen direction.
Key equation
potential gradient=Δ VΔ r
Work done by the electric field
Consider a positive test charge q moving a small distance Δr parallel to a uniform electric field E⃗. The electric force is F=qE in magnitude.
Key equation
F=qE
F
electric force (N)
q
test charge (C)
E
electric field intensity (N·C⁻¹)
Key equation
W=FΔ r
W
work done by the field for displacement along the force (J)
Key equation
W=qEΔ r
Work and potential energy
Electric potential energy changes by ΔU=qΔV. The work done by the electric field equals the negative change in electric potential energy.
Key equation
Δ U=qΔ V
\Delta U
change in electric potential energy (J)
Key equation
Wfield=-Δ U=-qΔ V
Equating the two expressions for work done by the field gives the field–potential relation.
Key equation
qEΔ r=-qΔ V
Key equation
E=-Δ VΔ r
Meaning of the negative sign
The negative sign is directional. Electric potential decreases when displacement is along E⃗ and increases when displacement is opposite to E⃗.
Motion of a positive charge
Work by electric field
Potential energy
Electric potential
Along E⃗
Positive
Decreases
Decreases
Opposite E⃗
Negative
Increases
Increases
Uniform and non-uniform fields
For a uniform field along one direction, the field and potential gradient are constant, so the finite-difference form applies directly.
Key equation
Ex=-Δ VΔ x
For a field that changes continuously with position, use the derivative at a point.
Key equation
Er=-dVdr
In three dimensions, the electric field is the negative gradient of the scalar potential.
Key equation
E=-∇ V
\nabla V
spatial rate and direction of greatest increase of potential (V·m⁻¹)
Parallel-plate magnitude
Between oppositely charged parallel plates, away from fringing edges, the field is approximately uniform. If the magnitude of the potential difference is V across plate separation d:
Key equation
E=|Δ V|d
d
perpendicular plate separation (m)
Unit equivalence: Assignment 10.3
Potential-gradient form gives V·m⁻¹, while force-per-charge form gives N·C⁻¹. They are equivalent because one volt is one joule per coulomb and one joule is one newton metre.
Key equation
1V=1JC,\qquad1J=1N·m
Key equation
1Vm=1JC·m=1N·mC·m=1NC
Textbook Example 10.3
At a point due to a point charge, electric field intensity is 24 N·C⁻¹ and electric potential is 6 J·C⁻¹. Find the distance of the point from the charge.
Key equation
E=24N·C-1,\qquad V=6J·C-1=6V
For the magnitude of the field and potential due to the same point charge, E=Vr. Therefore V=Er.
Key equation
r=VE=624m=0.25m
Exam-ready method
Define potential gradient as ΔVΔr.
Write F=qE and W=FΔr=qEΔr.
Write Wfield=−qΔV from potential-energy change.
Equate the work expressions and cancel q.
Obtain E=−ΔVΔr and explain the negative sign.
Show that V·m⁻¹=N·C⁻¹.
For Example 10.3, use r=VE and preserve the given values.
Did you understand this topic?
Your answer is saved with your learning record. It does not mark the topic as mastered.
Exam preparation
Exam Focus
Important points
Definition ΔVΔr.
Work-based derivation.
Negative-sign meaning.
Finite, differential and vector forms.
V/m=N/C proof.
Example 10.3 result 0.25 m.
Common mistakes
Omitting the negative sign in E=−ΔVΔr when discussing direction.
Interpreting the negative sign as a negative field magnitude.
Confusing work done by the field with work done by an external agent.
Using E=ΔVd instead of E=|ΔV|/d for a uniform field magnitude.
Writing V/m and N/C as different physical units.
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.
Potential Gradient
Sticky Notes — Quick Revision
1.
Key Point 1
Potential gradient is the change in potential per unit displacement, ΔVΔr.
2.
Key Point 2
The electric field is the negative potential gradient: E=−ΔVΔr for a uniform one-dimensional field.
3.
Key Point 3
The negative sign shows that electric potential decreases in the direction of E⃗.
4.
Key Point 4
For a point charge, E=Vr in magnitude, so textbook Example 10.3 gives r=VE=0.25 m.
Ready to test yourself?
You've completed the learning section.
8 concept questions
Easy → Medium → Hard
Explanation after each answer
Exam Practice
Important Question
Q1. Define potential gradient and derive its relation with electric field intensity.
Concept
Q2.What does the negative sign in E=−ΔVΔr mean?
Proof
Q3. Show that volt per metre is equivalent to newton per coulomb.
Numerical
Q4. Solve textbook Example 10.3.
Comparison
Q5. Distinguish the finite-difference and differential forms of the field–potential relation.
