Specification: AQA A-level Physics (7408) 3.7.4.2 | Paper: 2 | Data and formulae booklet: given in the exam
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Relative permittivity (dielectric constant), εᵣ: the ratio of the permittivity of a material to the permittivity of free space, εᵣ = ε ÷ ε₀. Equivalently, the ratio of the capacitance with the dielectric filling the gap to the capacitance with a vacuum between the plates. It has no unit (εᵣ = 1 for a vacuum, and about 1 for air).
Dielectric action: a dielectric is an insulator with polar molecules. In the electric field between the plates the polar molecules rotate and align with the field, producing their own electric field that opposes the applied field. This reduces the resultant field and the pd for the same charge, so more charge is stored per volt and the capacitance increases.
Parallel-plate capacitance: C is directly proportional to the overlapping plate area A and to εᵣ, and inversely proportional to the plate separation d. ε₀ = 8.85 × 10−12 F m⁻¹ is the permittivity of free space.
The equation
capacitance = (plate area × permittivity of free space × relative permittivity) ÷ plate separation C = Aε₀εᵣ ÷ d
| Quantity | Symbol | Unit |
|---|---|---|
| capacitance | C | farads (F) |
| area of overlap of the plates | A | square metres (m²) |
| permittivity of free space | ε₀ | farads per metre (F m⁻¹) |
| relative permittivity (dielectric constant) | εᵣ | no unit |
| separation of the plates | d | metres (m) |
Rearranged: d = Aε₀εᵣ ÷ C | A = Cd ÷ ε₀εᵣ | εᵣ = Cd ÷ Aε₀ | Convert cm² to m² by multiplying by 10−4 (1 cm² = 10−4 m²) and mm to m before you substitute.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Two parallel metal plates, each of area 0.010 m², are separated by 1.0 mm of air (εᵣ = 1.0). Calculate the capacitance in pF, to 3 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 1.0 mm = 1.0 × 10−3 m
F: C = Aε₀εᵣ ÷ d
I: C = 0.010 × 8.85 × 10−12 × 1.0 ÷ 1.0 × 10−3
F: C is already the subject
A: 8.85 × 10−11 F = 88.5 pF (3 s.f.)
Q2 (Core) A capacitor has plates of area 0.040 m² separated by a 0.50 mm sheet of polythene with relative permittivity 2.3. Calculate the capacitance in nF, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 0.50 mm = 5.0 × 10−4 m
F: C = Aε₀εᵣ ÷ d
I: C = 0.040 × 8.85 × 10−12 × 2.3 ÷ 5.0 × 10−4
F: C is already the subject
A: 1.628… × 10−9 F = 1.6 nF (2 s.f.)
Q3 (Core) An air-filled capacitor (εᵣ = 1.0) has plates of area 25 cm² and a capacitance of 50 pF. Calculate the plate separation in mm, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 25 cm² = 25 × 10−4 m² = 2.5 × 10−3 m²; 50 pF = 5.0 × 10−11 F
F: C = Aε₀εᵣ ÷ d
I: 5.0 × 10−11 = 2.5 × 10−3 × 8.85 × 10−12 × 1.0 ÷ d
F: d = 2.5 × 10−3 × 8.85 × 10−12 ÷ 5.0 × 10−11
A: 4.425 × 10−4 m = 0.44 mm (2 s.f.)
Watch out: 1 cm² = 10−4 m², not 10−2 m². Square the conversion factor as well as the number.
Q4 (Core) A capacitor is made from two circular plates of radius 6.0 cm, separated by 2.0 mm of air (εᵣ = 1.0). Calculate its capacitance in pF, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: r = 6.0 cm = 0.060 m, so A = πr² = π × 0.060² = 0.01130… m²; 2.0 mm = 2.0 × 10−3 m
F: C = Aε₀εᵣ ÷ d
I: C = 0.01130… × 8.85 × 10−12 × 1.0 ÷ 2.0 × 10−3
F: C is already the subject
A: 5.00… × 10−11 F = 50 pF (2 s.f.)
Q5 (Exam) An air-filled parallel-plate capacitor has a capacitance of 120 pF. (a) The plate separation is halved. State the new capacitance. (b) The gap is then completely filled with a dielectric of relative permittivity 4.0. Calculate the new capacitance and explain, in terms of the molecules of the dielectric, why it increases.
Show solution
Step 1 – (a) halving d
C ∝ 1 ÷ d, so halving d doubles C
A: C = 2 × 120 = 240 pF
Step 2 – (b) inserting the dielectric
C ∝ εᵣ, so C = 4.0 × 240
A: 960 pF
Explanation: the dielectric’s polar molecules rotate and align with the field between the plates. Their own field opposes the applied field, reducing the resultant field and so the pd for a given charge. More charge is stored per volt, so C increases.
Watch out: “the dielectric stores charge” earns no credit. The charge is stored on the plates; the dielectric only reduces the pd needed for that charge.
Q6 (Exam) A student measures the capacitance of a capacitor made from two plates of area 0.020 m² separated by a 0.10 mm sheet of an unknown insulator. The capacitance is 5.3 nF. Calculate the relative permittivity of the insulator, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 5.3 nF = 5.3 × 10−9 F; 0.10 mm = 1.0 × 10−4 m
F: C = Aε₀εᵣ ÷ d
I: 5.3 × 10−9 = 0.020 × 8.85 × 10−12 × εᵣ ÷ 1.0 × 10−4
F: εᵣ = Cd ÷ Aε₀ = (5.3 × 10−9 × 1.0 × 10−4) ÷ (0.020 × 8.85 × 10−12)
A: 2.99… = 3.0 (2 s.f., no unit)
Q7 (Exam) A capacitor is made from two rectangular plates, 15 cm × 20 cm, separated by a 0.25 mm sheet of mica (εᵣ = 6.0). It is connected to a 50 V supply. Calculate (a) the capacitance, (b) the charge stored and (c) the electric field strength between the plates, each to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: A = 0.15 × 0.20 = 0.030 m²; 0.25 mm = 2.5 × 10−4 m
Step 1 – (a) capacitance
F: C = Aε₀εᵣ ÷ d
I: C = 0.030 × 8.85 × 10−12 × 6.0 ÷ 2.5 × 10−4
A: 6.372 × 10−9 F = 6.4 nF (2 s.f.)
Step 2 – (b) charge
Q = CV = 6.372 × 10−9 × 50 = 3.186 × 10−7 C
A: 3.2 × 10−7 C (2 s.f.)
Step 3 – (c) field strength
E = V ÷ d = 50 ÷ 2.5 × 10−4
A: 2.0 × 105 V m⁻¹
Watch out: d must be in metres. Using 0.25 instead of 2.5 × 10−4 m makes C and E 1000 times too small.
Q8 (Exam) A computer keyboard key works as a capacitor. Each key has a plate of area 1.0 cm² above a fixed plate, with an air gap (εᵣ = 1.0) of 0.50 mm. Pressing the key reduces the gap to 0.20 mm. (a) Calculate the increase in capacitance in pF, to 2 significant figures. (b) The plates are held at a pd of 5.0 V. Calculate the extra charge that flows onto the plates when the key is pressed, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 1.0 cm² = 1.0 × 10−4 m²; 0.50 mm = 5.0 × 10−4 m; 0.20 mm = 2.0 × 10−4 m
Step 1 – capacitance before and after
F: C = Aε₀εᵣ ÷ d
I: before: C = 1.0 × 10−4 × 8.85 × 10−12 ÷ 5.0 × 10−4 = 1.77 × 10−12 F
after: C = 1.0 × 10−4 × 8.85 × 10−12 ÷ 2.0 × 10−4 = 4.425 × 10−12 F
A: increase = 4.425 − 1.77 = 2.655 pF = 2.7 pF (2 s.f.)
Step 2 – (b) extra charge
ΔQ = ΔC × V = 2.655 × 10−12 × 5.0 = 1.3275 × 10−11 C
A: 1.3 × 10−11 C (2 s.f.)
Q9 (Stretch) An air-filled capacitor (εᵣ = 1.0) has plates of area 0.020 m² separated by 2.0 mm. It is charged to 100 V. The plates are then pulled apart until the separation is 4.0 mm. Calculate the new charge, pd and electric field strength if (a) the capacitor stays connected to the 100 V supply and (b) the capacitor was disconnected from the supply before the plates were moved. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – starting values
C = Aε₀εᵣ ÷ d = 0.020 × 8.85 × 10−12 ÷ 2.0 × 10−3 = 8.85 × 10−11 F
Q = CV = 8.85 × 10−11 × 100 = 8.85 × 10−9 C; E = V ÷ d = 100 ÷ 2.0 × 10−3 = 50 000 V m⁻¹
Step 2 – new capacitance
d doubles, so C halves: C = 4.425 × 10−11 F
Step 3 – (a) still connected: V is fixed
V = 100 V; Q = CV = 4.425 × 10−11 × 100 = 4.425 × 10−9 C = 4.4 × 10−9 C (2 s.f.), so half the charge flows back to the supply
E = 100 ÷ 4.0 × 10−3 = 25 000 V m⁻¹ (halved)
Step 4 – (b) disconnected: Q is fixed
Q = 8.85 × 10−9 C (it has nowhere to go); V = Q ÷ C = 8.85 × 10−9 ÷ 4.425 × 10−11 = 200 V (doubled)
E = 200 ÷ 4.0 × 10−3 = 50 000 V m⁻¹ (unchanged)
Watch out: first decide what is held constant. Connected to the supply means V is constant; isolated means Q is constant.
Q10 (Stretch) An air-filled capacitor of capacitance 200 pF is charged to 12 V. A slab of dielectric with relative permittivity 2.5 is then slid in so that it completely fills the gap. Calculate the final charge and pd if (a) the capacitor stays connected to the 12 V supply and (b) the capacitor was disconnected before the slab was inserted. (c) Explain why the pd falls in (b).
Show solution
Step 1 – starting charge and new capacitance
Q = CV = 200 × 10−12 × 12 = 2.4 × 10−9 C
C ∝ εᵣ, so C = 2.5 × 200 = 500 pF = 5.0 × 10−10 F
Step 2 – (a) still connected: V is fixed
V = 12 V; Q = CV = 5.0 × 10−10 × 12 = 6.0 × 10−9 C, so an extra 3.6 × 10−9 C flows from the supply
Step 3 – (b) disconnected: Q is fixed
Q = 2.4 × 10−9 C; V = Q ÷ C = 2.4 × 10−9 ÷ 5.0 × 10−10 = 4.8 V
Step 4 – (c) explanation
the polar molecules of the dielectric align with the field, and their field opposes the field from the charges on the plates. The resultant field between the plates is reduced by a factor of εᵣ, and since V = Ed the pd falls by the same factor: 12 ÷ 2.5 = 4.8 V.
Watch out: the charge on an isolated capacitor cannot change, so any change in C must show up as a change in V.
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