Electric Field Strength of a Point Charge
Key idea: Define electric field strength, calculate the field due to point charges, determine direction, and apply vector superposition.
By the end, you can
- Define electric field strength and calculate the resultant field due to point charges.
1. Definitions (must know)
Electric field strength, vec E, at a point is the electric force per unit positive charge on a small stationary test charge placed at that point:
vec E = vec F/q
Electric field strength is a vector. Its direction is the direction of the force on a positive test charge. Its units are N C⁻¹, equivalent to V m⁻¹.
For a point source charge Q in free space or air, the field magnitude at distance r is
E = 1/4πε₀|Q|/r² = k|Q|/r²
The field points away from + Q and towards -Q.
2. Key ideas
- The source charge Q creates the field; a small test charge reveals it without significantly disturbing it.
- Epropto |Q| and Epropto 1/r² for a point charge.
- Use the sign of Q to decide field direction, not to make the field magnitude negative.
- The force on a separate charge q is vec F = qvec E; a negative q experiences force opposite to vec E.
- Fields obey superposition: add the individual field vectors at the point.
A field line shows the force direction for a positive test charge. A negative charge accelerates opposite to the field, and any initial sideways velocity can make the particle follow a curved path.
3. Detailed reasoning
A. From Coulomb force to field strength
Place a positive test charge q a distance r from source charge Q. Coulomb’s law gives
F = k|Qq|/r²
Dividing by the test charge magnitude gives
E = F/|q| = k|Q|/r²
The test charge cancels: the field is a property of the source configuration and position, not of the particular test charge used.
B. Superposition
For several source charges,
vec Eᵣₑₛᵤₗₜₐₙₜ = vec E₁ + vec E₂ + cdots
On a straight line, choose a positive direction and add signed components. In two dimensions, resolve each contribution into perpendicular components before adding.
4. Common mistakes
- Confusing E = kQ/r² with the potential relationship V = kQ/r.
- Giving only a magnitude when direction is required.
- Adding field magnitudes when the contributions point in opposite directions.
- Using the sign of a test charge to decide the field direction.
- Measuring r from the surface rather than from the point charge or centre of a spherical source model.
5. Exam tips
- Mark the point where E is required and draw a separate arrow for each source contribution.
- Write E = k|Q|/r² for magnitude, then state direction in words or with a signed component.
- Check inverse-square scaling: doubling r reduces E to one quarter.
- State the point-charge or spherically symmetric-source assumption when it matters.
6. Worked examples
Example 1: Field due to one point chargeCore
A charge Q = -6.0 nC is in free space. Find the electric field strength at a point 0.20 m away.
Show Answer
E = k|Q|/r² = (8.99 × 10⁹)(6.0 × 10⁻⁹)/(0.20)² = 1.35 × 10³ N C⁻¹
Because the source charge is negative, the field points towards Q.
Example 2: Resultant field between like chargesCore
Two identical positive charges are fixed at opposite ends of a line. What is the resultant field at the midpoint?
Show Answer
Each positive charge produces the same field magnitude at the midpoint, but the two fields point in opposite directions. Their vector sum is zero.
Example 3: From field to forceCore
An electron is placed in a field of strength 4.0 × 10⁴ N C⁻¹ directed east. Find the electric force on it.
Show Answer
|F| = |q|E = (1.60 × 10⁻¹⁹)(4.0 × 10⁴) = 6.4 × 10⁻¹⁵ N
The electron is negative, so its force is opposite to the field: west.
7. Mind stretchers
Mind stretcher 1: Zero field does not imply zero potentialExtension
At the midpoint between two identical positive charges, E = 0. Must the electric potential also be zero?
Show Answer
No. Field contributions are vectors and cancel, but potential contributions are scalars and add. Both potentials are positive, so the total potential is positive even though the resultant field is zero.
8. Practice, Quiz and Next Step
Close your notes and use Electric Field Strength of a Point Charge in the supplied context below. This requires a constructed explanation or working, not recognition of an option.
Fresh context: An unfamiliar data set or physical system requires you to apply Electric Field Strength of a Point Charge while stating the model, regime and assumptions.
- Retrieve: define electric field strength of a point charge in your own words, including units, sign or conditions where relevant.
- Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
- Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.
Check the response before looking back
- The model, regime, coordinates and assumptions are explicit.
- The derivation or multi-step reasoning is visible rather than implied.
- The conclusion is tested against units, data quality and a limiting case.
- A practical control, uncertainty or model limitation is evaluated where applicable.
If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.
Recommended next step
A Level Electric Fields Quiz
Why this will help: Use one focused question set to check that you can apply the lesson without prompts.
About 10 minutes