Relate electric potential to field strength
Define electric potential as work per unit positive charge and use the negative potential gradient to determine electric-field direction.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 18. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use force, charge, work, energy, potential difference, inverse-square relationships and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Electric potential at a point is defined as which quantity?
What you will learn
- Define electric potential using work per unit positive charge from infinity
- Relate field strength to the negative potential gradient
- Use potential change across a uniform field
Electric potential measures work per unit positive charge
Electric potential V at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. Its unit is J C⁻¹, called the volt. The zero of potential is taken at infinity for the point-charge model.
Potential is a scalar: it has magnitude and sign but no direction. The sign depends on the source charges and the chosen zero reference. Electric potential at a point is distinct from the potential energy of a particular test charge there.
Field points downhill in potential
The electric field at a point equals the negative of the potential gradient: along a chosen x direction, Eₓ = −dV/dx. The field points towards decreasing potential for a positive test charge. A steeper potential–distance graph means a stronger field magnitude.
For a uniform field, the magnitude is E = |ΔV|/Δd. If the potential increases with x, the field component Eₓ is negative; if potential decreases with x, Eₓ is positive. Keep this signed gradient separate from the magnitude formula.
Pause and explain
Potential increases as x increases. Which way is the electric field component?
Worked example
Potential rises from 200 V to 500 V as x increases by 0.015 m. Find the uniform field component Eₓ.
Show the worked solution
- Use Eₓ = −ΔV/Δx.
- ΔV = +300 V and Δx = +0.015 m.
- Interpret the negative sign as a field pointing towards decreasing x.
Answer Eₓ = −300/0.015 = −2.0 × 10⁴ V m⁻¹.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
The potential drops by 240 V over a distance of 0.012 m in the positive x direction. Find Eₓ and its magnitude.
Give me a hint
Use the signed relation Eₓ = −ΔV/Δx.
Compare my reasoning
- ΔV = −240 V and Δx = +0.012 m.
- Eₓ = −(−240)/(0.012) = +2.0 × 10⁴ V m⁻¹.
- The field points in the positive x direction and has magnitude 2.0 × 10⁴ V m⁻¹.
Eₓ = +2.0 × 10⁴ V m⁻¹; magnitude 2.0 × 10⁴ V m⁻¹.
Look for these in your work
- Signed potential change used
- Negative gradient applied
- Component and magnitude stated
A potential–distance graph is locally flat at a point. What does this indicate about the field component along that direction?
Give me a hint
Relate field to the gradient, not the potential value itself.
Compare my reasoning
- A locally flat graph has dV/dx = 0.
- Eₓ = −dV/dx.
- The field component along x is zero at that point in this one-dimensional interpretation.
The field component along x is zero where the potential gradient is zero.
Look for these in your work
- Gradient interpreted
- Field relation used
- Directional scope stated
At two locations, the potential–distance graph has gradients +100 V m⁻¹ and −300 V m⁻¹. Give the corresponding field components and compare magnitudes.
Give me a hint
Reverse the sign of each gradient to obtain Eₓ.
Compare my reasoning
- For gradient +100 V m⁻¹, Eₓ = −100 V m⁻¹.
- For gradient −300 V m⁻¹, Eₓ = +300 V m⁻¹.
- The second field magnitude is three times the first.
Eₓ values are −100 V m⁻¹ and +300 V m⁻¹; the second field is three times stronger.
Look for these in your work
- Both signs reversed
- Field components stated
- Magnitude ratio compared
Common mistakes
- Defining potential as energy rather than energy per unit positive charge
- Dropping the minus sign in the potential gradient
- Confusing signed field component with positive field magnitude
- Assuming potential and potential energy are the same quantity
What can you explain now?
Electric potential at a point is defined as which quantity?
Compare with the explanation
Work done per unit positive charge from infinity
This is the reference definition in the syllabus.
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Make it stick
Why does a negative potential gradient determine the electric-field direction?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.