- Define capacitance and use C = q / U and C = ε · ε₀ · S / d to find capacitance, charge and voltage
- Find the equivalent capacitance and the charge and voltage of each capacitor in series, parallel and mixed connections
- Calculate the energy of a capacitor and the energy density of the field
- Predict how C, q, U, E and W change when d, S or ε changes for a connected (U = const) or disconnected (q = const) capacitor
A camera flash shines for an instant, a defibrillator restarts a stopped heart with a strong electric pulse, and a smartphone screen feels the touch of your finger. All of these devices contain a capacitor — a component that stores charge and the energy of the electric field and releases them when needed. In the lesson «Electric charge, Coulomb’s law and the electric field» we studied charges and fields; now we learn how charge is “stored”. DİM tasks on this topic often hinge on one question: is the capacitor still connected to the source, or has it been disconnected?
Capacitance
When you give a conductor charge, its potential grows in direct proportion to the charge: double the charge and the voltage doubles too. So the ratio q / U is constant for a given conductor (or pair of conductors). In practice we use two conducting plates separated by a thin layer of dielectric — this is a capacitor. When it is connected to a source, one plate gets charge +q and the other −q. “The charge of a capacitor” means the magnitude of the charge on one plate, q: the total charge of the two plates is zero.
The ratio of the charge on one plate of a capacitor to the voltage between the plates; it shows how much charge the capacitor can store per volt. Its unit is the farad (F).
- Ccapacitance, F (farad)
- qmagnitude of the charge on one plate, C
- Uvoltage (potential difference) between the plates, V
1 F = 1 C/V. The farad is a very large unit, so µF (10⁻⁶ F), nF (10⁻⁹ F) and pF (10⁻¹² F) are used. Capacitance depends neither on the charge nor on the voltage: when q grows, U grows by the same factor and the ratio stays the same.
1) A 20 µF capacitor is connected to 50 V. Find its charge.
2) A capacitor stores 6 nC at 300 V. What is its capacitance?
3) The charge of a 5 µF capacitor is raised from 40 µC to 100 µC. How do the voltage and the capacitance change?
Show solutionHide solution
2) C = q / U = 6 · 10⁻⁹ / 300 = 2 · 10⁻¹¹ F = 20 pF.
3) U = q / C rises from 40 / 5 = 8 V to 100 / 5 = 20 V; the capacitance is still 5 µF — it depends on how the capacitor is built.
Capacitance of a parallel-plate capacitor
The simplest capacitor is the parallel-plate capacitor: two parallel plates of area S at a distance d from each other. Between the plates the field is uniform; outside it is almost zero. You can guess what the capacitance depends on: the wider the plates, the more charge “fits” on them (C ~ S); the closer the plates, the more strongly the positive and negative charges hold each other (C ~ 1/d).
When a dielectric is placed between the plates, it becomes polarised: the charges of its molecules weaken the field of the plates ε times. For the same charge q the voltage drops ε times, so the capacitance grows ε times. For vacuum and air ε ≈ 1, for paraffin about 2, for mica about 6 and for water 81.
- εrelative permittivity (dielectric constant) of the medium between the plates, no unit (1 for vacuum and air)
- ε₀electric constant, 8.85 · 10⁻¹² F/m
- Sarea of one plate, m²
- ddistance between the plates, m
The capacitance of a parallel-plate capacitor is directly proportional to the plate area and inversely proportional to the gap; it depends only on the geometry and the dielectric.
1) Each plate of a parallel-plate air capacitor has an area of 200 cm², and the gap is 1 mm. Find the capacitance.
2) The gap is completely filled with mica (ε = 6). What is the capacitance now?
3) If the plate area is doubled and the gap is tripled, how does the capacitance change?
Show solutionHide solution
C = ε₀ · S / d = 8.85 · 10⁻¹² · 0.02 / 10⁻³ = 1.77 · 10⁻¹⁰ F = 177 pF.
2) C′ = ε · C = 6 · 177 = 1062 pF ≈ 1.06 nF.
3) C′ / C = 2 / 3: the capacitance drops 1.5 times (the area doubles it, the gap divides it by 3).
The gap of a parallel-plate air capacitor is halved and completely filled with a substance of relative permittivity 3. How does the capacitance change?
A) increases 6 times B) increases 1.5 times C) decreases 6 times D) decreases 1.5 times E) does not change
Show solutionHide solution
Result: 3 · 2 = 6, the capacitance increases 6 times (A).
Option B (1.5) comes from dividing 3 by 2 — but d is in the denominator, so making it smaller increases C.
Capacitors in parallel and in series
Capacitors are combined to get the capacitance you need. In parallel, all capacitors are connected to the same two points: the voltage across each is the same and the charges add up — as if the plate area grew. In series, the capacitors form a chain: the source charges only the two outer plates, and the inner plates get the same charge q by induction. So every capacitor has the same charge and the voltages add up — as if the gap grew.
| Quantity | Parallel | Series |
|---|---|---|
| Voltage | U = U₁ = U₂ | U = U₁ + U₂ |
| Charge | q = q₁ + q₂ | q = q₁ = q₂ |
| Equivalent capacitance | C = C₁ + C₂ | 1/C = 1/C₁ + 1/C₂ |
| n equal capacitors (C₀) | C = n · C₀ | C = C₀ / n |
| Equivalent capacitance | larger than the largest one | smaller than the smallest one |
| Who gets more? | the larger capacitor — more charge | the smaller capacitor — more voltage |
- Cparequivalent capacitance in parallel: same voltage, charges add
- Cserequivalent capacitance in series: same charge, voltages add
For two capacitors in series C = C₁ · C₂ / (C₁ + C₂) — “product over sum”.
Capacitors C₁ = 2 µF and C₂ = 3 µF are connected to a 100 V source a) in parallel, b) in series. In each case find the equivalent capacitance and the charge and voltage of each capacitor. c) What is the equivalent capacitance of three 6 µF capacitors in series and in parallel?
Show solutionHide solution
b) C = 2 · 3 / (2 + 3) = 1.2 µF; both have the same charge: q = 1.2 · 100 = 120 µC.
U₁ = 120 / 2 = 60 V, U₂ = 120 / 3 = 40 V (check: 60 + 40 = 100). The smaller capacitor takes the larger voltage.
c) Series: C = 6 / 3 = 2 µF; parallel: C = 3 · 6 = 18 µF.
Note: µF · V = µC, so you can work without writing 10⁻⁶.
In the circuit, C₁ = 4 µF is in series with the parallel pair C₂ = 1 µF and C₃ = 3 µF; the source voltage is 12 V. What is the charge of C₃?
A) 6 µC B) 24 µC C) 18 µC D) 12 µC E) 36 µC
Show solutionHide solution
2) C₁ and the pair are in series: C = 4 · 4 / (4 + 4) = 2 µF.
3) Total charge (the charge of C₁ and of the pair): q = C · U = 2 · 12 = 24 µC.
4) Voltage across the pair: U₂₃ = q / C₂₃ = 24 / 4 = 6 V (C₁ also gets 6 V).
5) q₃ = C₃ · U₂₃ = 3 · 6 = 18 µC (C). Check: q₂ = 1 · 6 = 6 µC, 6 + 18 = 24 µC.
B (24) is the total charge; E (36) is the mistake of putting the whole 12 V across C₃.
Energy of a capacitor and energy density of the field
Connect a charged capacitor to a lamp and the lamp flashes — so the capacitor holds energy. This energy comes from the work the source did while charging it. During charging the voltage grows from 0 to U in proportion to the charge, so the average voltage at which the charge is moved is U / 2, which gives W = q · U / 2 (not q · U!). In other words, the energy is the area of the triangle under the U(q) graph.
- Wenergy of the capacitor (of its electric field), J
- qcharge of the capacitor, C
- Uvoltage between the plates, V
- Ccapacitance, F
The three forms are the same formula (substitute q = C · U). Which one to use depends on what stays constant: if U is constant, use C · U² / 2; if q is constant, use q² / (2C).
1) The 100 µF capacitor of a camera flash is charged to 300 V. Find its charge and energy.
2) A 40 µF capacitor carries 2 mC. What is its energy?
3) A defibrillator capacitor has a capacitance of 100 µF. To what voltage must it be charged to store 200 J?
Show solutionHide solution
2) Charge and capacitance are known: W = q² / (2C) = (2 · 10⁻³)² / (2 · 40 · 10⁻⁶) = 4 · 10⁻⁶ / (8 · 10⁻⁵) = 0.05 J = 50 mJ.
3) W = C · U² / 2 → U = √(2W / C) = √(2 · 200 / 10⁻⁴) = √(4 · 10⁶) = 2000 V = 2 kV.
Where is this energy stored? In the electric field between the plates. In a parallel-plate capacitor the field is uniform and its strength is E = U / d. Substituting C = ε · ε₀ · S / d and U = E · d into W = C · U² / 2 gives W = (ε · ε₀ · E² / 2) · S · d, and S · d is the volume filled by the field. So the energy per unit volume — the energy density — depends only on the field strength and the medium.
- Estrength of the uniform field between the plates, V/m
- ddistance between the plates, m
- wenergy density of the electric field, J/m³; the capacitor’s energy is W = w · S · d
The energy density grows as the square of the field strength: double the field and every cubic metre holds four times the energy.
A parallel-plate air capacitor has plates of area 100 cm², a gap of 2 mm and a voltage of 200 V.
1) Find the field strength between the plates and the energy density.
2) Calculate the capacitor’s energy in two ways: w · S · d and C · U² / 2.
Show solutionHide solution
2) Volume S · d = 0.01 · 0.002 = 2 · 10⁻⁵ m³; W = 0.04425 · 2 · 10⁻⁵ ≈ 8.85 · 10⁻⁷ J.
C = ε₀ · S / d = 8.85 · 10⁻¹² · 0.01 / 0.002 = 4.425 · 10⁻¹¹ F; W = C · U² / 2 = 4.425 · 10⁻¹¹ · 4 · 10⁴ / 2 = 8.85 · 10⁻⁷ J — the same result.
A 1 µF capacitor is charged to 300 V, disconnected from the source and then connected in parallel to an uncharged 2 µF capacitor. Calculate the common voltage across the capacitors (in V).
Show solutionHide solution
After joining they are in parallel: C = 1 + 2 = 3 µF, with one common voltage:
U = q / C = 300 / 3 = 100. Answer: 100.
The energy, however, is not conserved: by C · U² / 2 it drops from 45 mJ to 15 mJ, and 30 mJ becomes heat in the wires.
Connected to the source or disconnected?
DİM’s favourite capacitor question goes like this: the plates are moved apart (or together), or a dielectric is inserted — how do C, q, U, E and W change? The key is to find what stays constant. If the capacitor stays connected to the source, the source keeps the voltage: U = const. If it was charged and then disconnected, the charge has nowhere to go: q = const.
- 1Pick the constant
Connected → U = const; disconnected → q = const.
- 2Find the capacitance
C = ε · ε₀ · S / d: C grows when d decreases, S increases or a dielectric is inserted.
- 3Find the second quantity
If U = const, q = C · U changes together with C; if q = const, U = q / C changes opposite to C.
- 4Find the field
E = U / d. If q = const, E = q / (ε · ε₀ · S) — it does not depend on the gap!
- 5Find the energy
U = const: W = C · U² / 2 changes like C; q = const: W = q² / (2C) changes opposite to C.
| Quantity | d decreases n times, U = const | d decreases n times, q = const | dielectric (ε), U = const | dielectric (ε), q = const |
|---|---|---|---|---|
| C | × n | × n | × ε | × ε |
| q | × n | unchanged | × ε | unchanged |
| U | unchanged | ÷ n | unchanged | ÷ ε |
| E | × n | unchanged | unchanged | ÷ ε |
| W | × n | ÷ n | × ε | ÷ ε |
Two identical parallel-plate air capacitors are charged to the same voltage. The first stays connected to a constant-voltage source; the second is disconnected. The gaps of both are then completely filled with a dielectric. In the Euler–Venn diagram, I holds the statements true only for the connected capacitor, II only for the disconnected one, and III for both:
1. The capacitance increases.
2. The charge does not change.
3. The voltage between the plates decreases.
4. The field strength between the plates does not change.
5. The energy increases.
A) I – 2, 4; II – 3, 5; III – 1
B) I – 5; II – 2, 3, 4; III – 1
C) I – 1, 4; II – 2, 3; III – 5
D) I – 4, 5; II – 2, 3; III – 1
E) I – 4, 5; II – 3; III – 1, 2
Show solutionHide solution
Disconnected (q = const): C grows; q is unchanged; U = q / C drops; E = U / d drops; W = q² / (2C) drops → true statements 1, 2, 3.
The only common statement is 1 (III). Only the connected one: 4, 5 (I); only the disconnected one: 2, 3 (II). Answer: D.
1) A parallel-plate capacitor stays connected to a constant-voltage source, and the gap is increased 3 times. How do the capacitance, charge, field strength and energy change? Justify with formulas.
2) A charged capacitor has been disconnected from the source; its energy is 6 mJ. The gap is increased 3 times. Calculate the new energy (in mJ).
Show solutionHide solution
C = ε · ε₀ · S / d → d triples, so C drops 3 times.
q = C · U → drops 3 times (part of the charge flows back to the source).
E = U / d → drops 3 times.
W = C · U² / 2 → drops 3 times.
2) q = const, W = q² / (2C). C drops 3 times → W triples: W = 3 · 6 = 18. Answer: 18.
The extra energy is the work of the external force that pulls the attracting plates apart.
Key points
- Capacitance C = q / U, unit farad (F); it does not depend on the charge or the voltage.
- Parallel-plate capacitor: C = ε · ε₀ · S / d; a dielectric multiplies the capacitance by ε.
- Parallel: same U, C = C₁ + C₂; series: same q, 1/C = 1/C₁ + 1/C₂.
- Energy W = q · U / 2 = C · U² / 2 = q² / (2C); energy density w = ε · ε₀ · E² / 2, with E = U / d in a uniform field.
- A connected capacitor keeps U = const, a disconnected one keeps q = const — everything else follows from that.
Check yourself
12 questions. Every correct answer earns XP.