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.
Potential, potential energy and negative gradient 14(c)–(f)
Question 1
At 0.30 m from a −6.0 nC point charge, find V and the potential energy of a +2.0 nC charge. State the field direction from the potential gradient.
Check the model response
V = kQ/r = −180 V. Uᴱ = qV = kQq/r = −3.60 × 10⁻⁷ J. Since E is the negative potential gradient, the field points towards decreasing potential: radially towards the negative source.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Potential, potential energy and negative gradient 14(c)–(f)
Check this idea
Misconception: Potential contributions must be added as vectors.
Repair: Potential is scalar, so contributions add algebraically; field contributions add as vectors.
Check this idea
Misconception: Electric field points towards increasing potential.
Repair: E is the negative potential gradient and points towards decreasing potential.
Check this idea
Misconception: Work done by the field is qΔV.
Repair: ΔU = qΔV, while work done by the electric field is −ΔU.
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.
Potential, potential energy and negative gradient 14(c)–(f)
Model 1
For a +4.0 nC source charge, compare V and E at 0.20 m and 0.40 m, then find Uᴱ for a −3.0 nC test charge at 0.20 m.
Check the model response
V = kQ/r, so doubling r halves V: 180 V to 89.9 V. E = kQ/r², so it quarters: 899 N C⁻¹ to 225 N C⁻¹. Uᴱ = qV = (−3.0 × 10⁻⁹)(180) = −5.39 × 10⁻⁷ J.
guided practice
4. Guided practice
Use each hint only to select the inverse-square relation, sign convention, gradient, force direction or energy form.
Potential, potential energy and negative gradient 14(c)–(f)
Question 1
Potential falls linearly from 120 V to 40 V over +0.20 m. Find the signed field component.
Hint: Calculate the signed potential gradient before applying the minus sign.
Check the model response
dV/dx = (40 − 120)/0.20 = −400 V m⁻¹, so Eₓ = −dV/dx = +400 V m⁻¹.
independent practice
5. Independent practice
Solve without repair notes and state every point-charge, free-space, uniform-field and non-relativistic assumption used.
Potential, potential energy and negative gradient 14(c)–(f)
Question 1
Define electric potential using external work, then relate V, Uᴱ and E for a point charge.
Check the model response
V is external work per unit charge in bringing a small positive test charge slowly from infinity to the point. For source Q, V = Q/(4πε₀r); a two-charge system has Uᴱ = qV = Qq/(4πε₀r). Locally, E = −dV/dr, so the field points down the potential gradient.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Potential, potential energy and negative gradient 14(c)–(f)
Question 1
Define V, then find V and Uᴱ at 0.25 m from +6.0 nC for a −2.0 nC test charge. State the field direction from E = −dV/dr.
Check the model response
V is external work per unit positive charge from infinity. V = 216 V and Uᴱ = qV = −4.32 × 10⁻⁷ J. V decreases with increasing r, so −dV/dr is positive radial: the field points outward.
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.
Potential, potential energy and negative gradient 14(c)–(f)
Question 1
A +3.0 nC charge moves through ΔV = −50 V. Find ΔU and work done by the field.
Check the model response
ΔU = qΔV = −1.50 × 10⁻⁷ J. Work done by the field is −ΔU = +1.50 × 10⁻⁷ J.