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.
Capacitance and stored electric potential energy 14(j)–(k)
Question 1
A 47 μF capacitor is charged to 12 V. Find its charge and stored energy, and identify the relevant graph area.
Check the model response
Q = CV = 5.64 × 10⁻⁴ C. U = ½QV = ½CV² = 3.38 × 10⁻³ J. It is the triangular area under a graph of potential difference V against charge Q.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Capacitance and stored electric potential energy 14(j)–(k)
Check this idea
Misconception: A capacitor stores charge on only one plate.
Repair: Equal and opposite charges reside on its two plates; Q denotes the magnitude on either plate.
Check this idea
Misconception: Stored energy is QV because every increment of charge crosses the final p.d.
Repair: The p.d. rises during charging, so energy is the triangular V–Q area ½QV for constant capacitance.
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.
Capacitance and stored electric potential energy 14(j)–(k)
Model 1
A capacitor stores 18 mJ at 30 V. Find C and Q using two equivalent energy equations.
Check the model response
From U = ½CV², C = 2U/V² = 4.0 × 10⁻⁵ F = 40 μF. Then Q = CV = 1.20 × 10⁻³ C. Check: ½QV = 18 mJ.
guided practice
4. Guided practice
Use each hint only to select the inverse-square relation, sign convention, gradient, force direction or energy form.
Capacitance and stored electric potential energy 14(j)–(k)
Question 1
A 20 μF capacitor holds 0.50 mC. Find V and U.
Hint: Convert both prefixes before using C = Q/V.
Check the model response
V = Q/C = 25 V. U = ½Q²/C = 6.25 × 10⁻³ J.
independent practice
5. Independent practice
Solve without repair notes and state every point-charge, free-space, uniform-field and non-relativistic assumption used.
Capacitance and stored electric potential energy 14(j)–(k)
Question 1
Derive the three capacitor-energy forms from the V–Q graph and C = Q/V.
Check the model response
For constant C, the V–Q graph is a straight line from the origin, so its area is U = ½QV. Substituting V = Q/C gives U = ½Q²/C; substituting Q = CV gives U = ½CV².
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Capacitance and stored electric potential energy 14(j)–(k)
Question 1
A 15 μF capacitor is at 20 V. Calculate Q and U, and explain the factor ½.
Check the model response
Q = CV = 3.0 × 10⁻⁴ C. U = ½CV² = 3.0 × 10⁻³ J. The charging p.d. grows linearly from zero to V, so the area under the V–Q graph is triangular.
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.
Capacitance and stored electric potential energy 14(j)–(k)
Question 1
At fixed capacitance, voltage doubles. State the charge and energy factors.
Check the model response
Q = CV doubles. U = ½CV² increases by a factor of four.