Electric field strength due to a point charge
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 does a source charge shape the field around it?
Electric field strength is force per unit positive test charge. A point source gives E = Q/(4πε₀r²), directed away from positive Q and toward negative Q. Field-line density represents strength, while vector superposition locates neutral points or resultant directions.
Define field independently of the test charge
Electric field strength is force per unit positive test charge: E = F/q. A point source Q produces E = Q/(4πε₀r²) in magnitude, directed away from positive Q and towards negative Q.
The source determines E; a test charge then experiences F = qE. A negative test charge feels force opposite to the field. Keeping these two steps separate avoids inserting the test charge into the point-field equation.
Check your understanding: A negative charge is released in a field pointing east. Which way is its initial acceleration?
West, because F = qE reverses direction for q < 0.
Read field-line models and neutral points
Field lines show direction by their tangent and relative strength by their density. They begin on positive charge and end on negative charge or at infinity, never cross, and are not physical paths that charges must follow.
Fields from several sources add vectorially. A neutral point has zero resultant field, but its potential need not be zero because potential adds as a scalar. Between unequal like charges, the neutral point lies closer to the smaller charge.
Check your understanding: Why can field lines not cross?
A field vector at one point has one unique direction; crossing lines would assign two directions.
Key ideas to keep
- Field direction is defined for a positive test charge.
- The test charge does not appear in the field expression.
- At a neutral point, fields cancel; potentials need not be zero.
See the reasoning
Worked example
Add two point-charge fields at the midpoint
Question: Charges +8.0 nC and −2.0 nC lie 0.30 m apart. Find the resultant field at the midpoint.
Step 1: Draw both field arrows
Why: The + source points away and the − source points towards itself.
Working: At the midpoint both contributions point from +8.0 nC towards −2.0 nC.
Step 2: Calculate each magnitude
Why: Each source is 0.15 m from the midpoint.
Working: E₊ = 3.20×10³ N C⁻¹ and E₋ = 799 N C⁻¹.
Step 3: Add vectors
Why: The two arrows have the same direction here.
Working: E = 4.00×10³ N C⁻¹ towards the negative charge.
Answer: At the midpoint r = 0.15 m. The + charge produces 3.20 × 10³ N C⁻¹ away from itself; the − charge produces 799 N C⁻¹ towards itself. Both directions are from + towards −, so E = 4.00 × 10³ N C⁻¹ towards the negative charge.
Check: Do not cancel the fields merely because the source signs differ; their directions at the chosen point decide.
Use a hint if needed
Practise with support
Try this
A negative point charge produces field magnitude 500 N C⁻¹ at 0.10 m. Find the magnitude at 0.25 m and state the direction.
Hint: Use the distance ratio squared; direction follows the source sign.
Check your answer
E₂ = 500(0.10/0.25)² = 80 N C⁻¹. The field points radially towards the negative charge.
Now work without the hint
Practise independently
Your turn
Derive the field expression for a point charge from Coulomb force and the field definition.
Check your answer
Field strength is force per unit positive test charge: E = F/q. Substituting F = Qq/(4πε₀r²) gives E = Q/(4πε₀r²). Direction is away from positive Q and towards negative Q.
Avoid these traps
Common mistakes
Common mistake
Field direction depends on the sign of the test charge.
What is wrong with this reasoning?
Show better thinking
Field direction is defined by force on a positive test charge and depends on the source charges.
Write for the examiner
Exam guidance
Keep field and force separate: use the source to find E, then the test charge to find F = qE.
Exam-style practice [4 marks]
Find the field magnitude and direction 0.50 m from a −10 nC point charge.
Plan before you answer
- Use the source charge only in E = kQ/r².
- Calculate the magnitude.
- Give the radial direction.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
E = (8.99 × 10⁹)(10 × 10⁻⁹)/(0.50)² = 360 N C⁻¹, directed towards the negative charge.
Come back in three days
Check what stayed with you
Recall question
A +2.0 nC point charge produces a field at 0.30 m. Find its magnitude.
Check the answer
E = (8.99 × 10⁹)(2.0 × 10⁻⁹)/(0.30)² = 200 N C⁻¹, radially outward.
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(b) · 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