Electric potential & capacitance
The idea
Electric potential is potential energy per unit charge, and it converts vector field problems into scalar bookkeeping: the field is the negative gradient of V, so field lines run downhill in potential, and the potential difference between two points is the negative integral of E along any path between them. For the uniform field between parallel plates this collapses to the workhorse relation E = V/d. You already used voltage in circuits; here it acquires its field-theoretic meaning.
A capacitor stores charge and energy by holding +Q and −Q apart: C = Q/V measures how much charge per volt the geometry can hold, and for parallel plates C = ε₀A/d — bigger plates and smaller gaps store more. The stored energy U = ½CV² = Q²/(2C) lives in the field itself, which is why a dielectric slab (which weakens the internal field) raises capacitance.
Beware the reflex that V = 0 implies E = 0 or vice versa. The potential is a running total of the field along a path, so the field can be strong where the potential happens to pass through zero (midway between equal and opposite charges), and the potential can be large and constant where the field vanishes (inside a charged conducting shell). Field is slope; potential is height.
Worked example
A parallel-plate capacitor has plate area 2.0 × 10⁻² m² and separation 1.0 mm, with vacuum between the plates. It is charged to 12 V. Using ε₀ = 8.85 × 10⁻¹² F/m, find the capacitance, the stored charge, the field between the plates, and the stored energy.
- Compute the capacitance from geometry: C = ε₀A/d = (8.85 × 10⁻¹² × 2.0 × 10⁻²)/(1.0 × 10⁻³) = 1.77 × 10⁻¹⁰ F = 177 pF.
- Find the charge from the definition of capacitance: Q = CV = 1.77 × 10⁻¹⁰ × 12 ≈ 2.1 × 10⁻⁹ C, i.e. +2.1 nC on one plate and −2.1 nC on the other.
- Get the field from the uniform-field relation: E = V/d = 12/(1.0 × 10⁻³) = 1.2 × 10⁴ V/m, directed from the positive plate to the negative plate.
- Compute the stored energy: U = ½CV² = 0.5 × 1.77 × 10⁻¹⁰ × 12² ≈ 1.3 × 10⁻⁸ J.
- Sanity-check the scale: tens of picofarads and tens of nanojoules are typical for bench-top plates — capacitors store charge readily but only modest energy, which is why farad-scale supercapacitors need enormous effective areas.
Answer. C ≈ 177 pF, Q ≈ 2.1 nC, E = 1.2 × 10⁴ V/m, and the stored energy is about 1.3 × 10⁻⁸ J.
Check your understanding
- If the battery is disconnected and the plates are pulled farther apart, which of Q, V, E, and U change, and why?
- Why does inserting a dielectric increase capacitance — what does the material do to the field between the plates?
- How can the potential be zero at a point where the electric field is strong — what picture reconciles them?
- Why does the energy formula have three equivalent forms (½CV², ½QV, Q²/2C), and when is each most convenient?
Build the foundations first
Electric potential & capacitance builds on these concepts. If any feel shaky, start there.