Electric fields: force, potential and capacitors
Key idea: Keep vector force and field distinct from scalar potential and energy, then use the negative potential gradient, uniform fields and capacitor graphs with explicit signs and assumptions.
Before you start: Energy & Fields objective chainMotion & Forces objective chain
By the end, you can
- Apply Coulomb's law and the point-charge field equation with correct inverse-square scaling and direction.
- Define and calculate electric potential and two-charge potential energy, then use the negative potential gradient.
- Analyse force, acceleration and trajectory in a uniform electric field.
- Define capacitance and obtain stored-energy equations from the area under a potential-difference–charge graph.
Starting-point self-check
1. Check your starting point
Attempt all five groups without notes and mark the first inverse-square, sign, scalar/vector, gradient or graph-area decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Electric field strength due to a point charge 14(b)
Question 1
Find the electric field strength 0.20 m from a +4.0 nC point charge in air, including direction.
Check the model response
E = Q/(4πε₀r²) = (8.99 × 10⁹)(4.0 × 10⁻⁹)/(0.20)² = 899 N C⁻¹. It points radially away from the positive source charge.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Electric field strength due to a point charge 14(b)
Check this idea
Misconception: Field direction depends on the sign of the test charge.
Repair: Field direction is defined by force on a positive test charge and depends on the source charges.
worked example
3. Follow five worked models
Follow how each solution fixes the source geometry, vector direction, potential reference, field uniformity or graph axes before calculating.
Electric field strength due to a point charge 14(b)
Model 1
Charges +8.0 nC and −2.0 nC lie 0.30 m apart. Find the resultant field at the midpoint.
Check the model response
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.
guided practice
4. Guided practice
Use each hint only to select the inverse-square relation, sign convention, gradient, force direction or energy form.
Electric field strength due to a point charge 14(b)
Question 1
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 the model response
E₂ = 500(0.10/0.25)² = 80 N C⁻¹. The field points radially towards the negative charge.
independent practice
5. Independent practice
Solve without repair notes and state every point-charge, free-space, uniform-field and non-relativistic assumption used.
Electric field strength due to a point charge 14(b)
Question 1
Derive the field expression for a point charge from Coulomb force and the field definition.
Check the model response
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.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Electric field strength due to a point charge 14(b)
Question 1
Find the field magnitude and direction 0.50 m from a −10 nC point charge.
Check the model response
E = (8.99 × 10⁹)(10 × 10⁻⁹)/(0.50)² = 360 N C⁻¹, directed towards the negative charge.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Electric field strength due to a point charge 14(b)
Question 1
A +2.0 nC point charge produces a field at 0.30 m. Find its magnitude.
Check the model response
E = (8.99 × 10⁹)(2.0 × 10⁻⁹)/(0.30)² = 200 N C⁻¹, radially outward.