Topic 10.413 min read
Potential Gradient
Potential gradient is the rate of change of electric potential with displacement. The electric field points in the direction of decreasing potential, so in one dimension E=−ΔVΔr for a uniform field and E=−dV/dr at a point. Its unit, volt per metre, is equivalent to newton per coulomb.
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The main idea
Potential gradient tells how quickly electric potential changes as we move through an electric field. A large change over a short distance means a strong electric field.
- E
- electric field component along the displacement (V·m⁻¹)
- \Delta V
- change in electric potential (V)
- \Delta r
- displacement (m)
Meaning of the negative sign
- Move in the direction of E⃗: electric potential decreases.
- Move opposite to E⃗: electric potential increases.
- A positive charge released freely moves toward lower potential.
- Volt per metre and newton per coulomb are the same field unit.
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Potential difference and displacement
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.
- 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.
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.
- F
- electric force (N)
- q
- test charge (C)
- E
- electric field intensity (N·C⁻¹)
- W
- work done by the field for displacement along the force (J)
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.
- \Delta U
- change in electric potential energy (J)
Equating the two expressions for work done by the field gives the field–potential relation.
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.
For a field that changes continuously with position, use the derivative at a point.
In three dimensions, the electric field is the negative gradient of the scalar potential.
- \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:
- 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.
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.
For the magnitude of the field and potential due to the same point charge, E=Vr. Therefore V=Er.
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.
Exam Focus
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.
Important formulas
- E=-Δ VΔ r
- E=-dVdr
- E=-∇ V
- W_{field}=-qΔ V
Common exam traps
- Use Wfield=−ΔU; external work has the opposite sign for a slow move.
- Use magnitudes only when writing E=|ΔV|/d.
Long question structure
Definition → W=FΔr → F=qE → Wfield=−qΔV → equate → E=−ΔVΔr → sign → units → example
Common Mistakes
Common Mistakes
- Mistake
Omitting the negative sign in E=−ΔVΔr when discussing direction.
- Mistake
Interpreting the negative sign as a negative field magnitude.
- Mistake
Confusing work done by the field with work done by an external agent.
- Mistake
Using E=ΔVd instead of E=|ΔV|/d for a uniform field magnitude.
- Mistake
Writing V/m and N/C as different physical units.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Potential difference is ΔV=VB−VA between two points A and B.
Potential gradient is the change in potential per unit displacement, ΔVΔr.
The electric field is the negative potential gradient: E=−ΔVΔr for a uniform one-dimensional field.
In differential form, E=−dV/dr; generally, E⃗=−∇V.
Potential gradient is the change in electric potential per unit displacement; electric field is its negative spatial gradient.
E=-Δ VΔ r, E=-dVdr, E=-∇ V
- Definition
- Potential gradient is the change in electric potential per unit displacement; electric field is its negative spatial gradient.
- Formula
- E=-Δ VΔ r, E=-dVdr, E=-∇ V
- Example
- For V=6 V and E=24 V·m⁻¹, r=VE=0.25 m.
- Critical warning
- The minus sign gives direction toward decreasing potential; field magnitude remains non-negative.
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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.