Parallel plate capacitor, with and without a dielectric
A parallel plate capacitor in vacuum has C = ε₀A/d. Filling the gap with a dielectric of constant K makes it K times larger: C = Kε₀A/d.
Derivation: capacitance without a dielectric
Two large parallel plates, each of area A, are a distance d apart in vacuum. They carry +Q and −Q.
- 1Surface charge density σ = Q/A. Since d is small compared with the plate size, ignore the edge (fringing) effects.
- 2Each plate acts like an infinite sheet, with field σ/(2ε₀).
- 3Between the plates, both fields point from + to −, so they add: E = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀.
- 4Outside the plates, the two fields are opposite, so they cancel: E = 0.
- 5Potential difference: V = Ed = σd/ε₀ = Qd/(ε₀A).
- 6★So, C₀ = Q/V = ε₀A/d.
★ key step
Derivation: capacitance with a dielectric filling the gap
Now fill the whole gap with a dielectric of constant K, keeping the charges ±Q on the plates.
- 1The dielectric is polarised. Charges −σₚ and +σₚ appear on its faces, next to the + and − plates.
- 2The net field inside is E = (σ − σₚ)/ε₀, which is smaller than σ/ε₀.
- 3Write σ − σₚ = σ/K. So E = σ/(Kε₀) = E₀/K.
- 4Potential difference: V = Ed = σd/(Kε₀) = Qd/(Kε₀A).
- 5★So, C = Q/V = Kε₀A/d = KC₀. The capacitance rises K times.
★ key step
How to draw the figure
- Draw two vertical parallel lines for the plates. Label the gap d and the plate area A.
- Write + signs along the left plate (+Q) and − signs along the right plate (−Q).
- Draw equally spaced arrows from left to right between the plates for the uniform field E. Draw no arrows outside.
- With a dielectric: shade a slab between the plates, put − signs on its left face and + signs on its right face, draw fewer arrows inside and label E = E₀/K.
Worked example: capacitance and charge
Two plates, each 20 cm × 10 cm, are 1 mm apart in air. Find (a) the capacitance, (b) the charge when it is joined to a 50 V supply and (c) the capacitance when the gap is filled with a dielectric of K = 5. (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
- A = 0.20 m × 0.10 m = 2 × 10⁻² m²; d = 1 × 10⁻³ m.
- (a) C₀ = ε₀A/d = 8.85 × 10⁻¹² × 2 × 10⁻² ÷ (1 × 10⁻³) = 1.77 × 10⁻¹⁰ F = 177 pF.
- (b) Q = C₀V = 1.77 × 10⁻¹⁰ × 50 = 8.85 × 10⁻⁹ C.
- (c) C = KC₀ = 5 × 177 pF = 885 pF.
(a) 177 pF; (b) 8.85 × 10⁻⁹ C (about 8.9 nC); (c) 885 pF
Worked example: a slab that fills half the gap
A parallel plate capacitor has plate separation 4 mm and capacitance C₀ in air. A dielectric slab of thickness 2 mm and dielectric constant 4, with the same area as the plates, is placed between them. Find the new capacitance in terms of C₀.
- Keep the charge Q fixed. In air the field is E₀, so without the slab V₀ = E₀d = E₀ × 4 mm.
- With the slab, the field is E₀ in the 2 mm of air and E₀/K = E₀/4 in the 2 mm of slab.
- V = E₀ × 2 mm + (E₀/4) × 2 mm = E₀ × 2.5 mm.
- C = Q/V and Q is unchanged, so C/C₀ = V₀/V = 4 ÷ 2.5 = 1.6.
- Extra help, beyond your textbook: the same method gives C = ε₀A/(d − t + t/K) for a slab of any thickness t.
C = 1.6 C₀: larger than C₀, but less than 4C₀, because the slab fills only half the gap
What changes C and what does not
Your phone's touchscreen is a grid of tiny capacitors. Your finger, which holds salty water, changes the capacitance where you touch, and the phone's chip works out exactly where you tapped.
3 marks: “Derive an expression for the capacitance of a parallel plate capacitor with vacuum between its plates.” Field between the plates E = σ/ε₀; V = Ed = Qd/(ε₀A); C = Q/V = ε₀A/d.
2 marks: “A parallel plate capacitor has capacitance C. Its plate separation is doubled and the gap is filled with a dielectric of K = 4. Find the new capacitance.” C′ = Kε₀A/(2d) = (4/2)C = 2C.
5 marks: “Derive the capacitance of a parallel plate capacitor filled with a dielectric of constant K. Explain why the capacitance increases.” Polarisation charges ±σₚ reduce the field to σ/(Kε₀); V = Qd/(Kε₀A); C = Kε₀A/d. The dielectric lowers V for the same Q, so C = Q/V rises.
C = ε₀A/d: area up, distance down, dielectric K times.