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.
Uniform fields, force and charged-particle motion 14(g)–(i)
Question 1
Parallel plates are 12 mm apart with a p.d. of 360 V. Find E, then the force and acceleration of a proton. Use e = 1.60 × 10⁻¹⁹ C and mp = 1.67 × 10⁻²⁷ kg.
Check the model response
E = ΔV/d = 3.00 × 10⁴ V m⁻¹. F = qE = 4.80 × 10⁻¹⁵ N along the field. a = F/m = 2.87 × 10¹² m s⁻² while the field is uniform.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Uniform fields, force and charged-particle motion 14(g)–(i)
Check this idea
Misconception: E = ΔV/d applies to every field.
Repair: It applies directly only across a uniform field; a non-uniform field requires a local gradient.
Check this idea
Misconception: A charged particle must follow an electric field line.
Repair: The field fixes acceleration; an initial transverse velocity produces a curved path.
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.
Uniform fields, force and charged-particle motion 14(g)–(i)
Model 1
An electron enters horizontally at 3.0 × 10⁷ m s⁻¹ between horizontal plates where E = 2.0 × 10⁴ N C⁻¹ downward. Find its vertical acceleration and describe its path.
Check the model response
The electron force is opposite the field, upward. Its magnitude is eE = 3.20 × 10⁻¹⁵ N, so a = 3.51 × 10¹⁵ m s⁻² upward. Horizontal velocity stays constant while vertical velocity changes uniformly, producing a parabolic path within the uniform field.
guided practice
4. Guided practice
Use each hint only to select the inverse-square relation, sign convention, gradient, force direction or energy form.
Uniform fields, force and charged-particle motion 14(g)–(i)
Question 1
A −3.0 nC charge is between plates with E = 2.0 × 10⁵ N C⁻¹ to the right. Find its force.
Hint: The field is defined for a positive test charge; retain the charge sign.
Check the model response
F = qE = (−3.0 × 10⁻⁹)(2.0 × 10⁵) = −6.0 × 10⁻⁴ N, meaning 6.0 × 10⁻⁴ N to the left.
independent practice
5. Independent practice
Solve without repair notes and state every point-charge, free-space, uniform-field and non-relativistic assumption used.
Uniform fields, force and charged-particle motion 14(g)–(i)
Question 1
A positive particle enters a uniform field perpendicular to its initial velocity. Explain its motion and how the answer changes for a negative particle.
Check the model response
The constant force qE gives constant acceleration parallel to E for positive q, while the perpendicular velocity component stays constant, producing a parabola. For negative q the acceleration and curvature reverse; the path need not follow a field line.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Uniform fields, force and charged-particle motion 14(g)–(i)
Question 1
Plates 8.0 mm apart have 240 V across them. Find E and the force on an electron, then describe its acceleration direction.
Check the model response
E = 240/0.0080 = 3.0 × 10⁴ V m⁻¹. Force magnitude is eE = 4.8 × 10⁻¹⁵ N, opposite E because the electron is negative; its acceleration has the same direction as its force.
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.
Uniform fields, force and charged-particle motion 14(g)–(i)
Question 1
A charged particle starts from rest in a uniform field. State the form of its motion and one condition under which that description ends.
Check the model response
It has constant acceleration a = qE/m in a straight line parallel or antiparallel to E. The description ends on leaving the uniform region or when relativistic effects, collisions or another force becomes significant.