Cambridge AS & A Level9702

Capacitance

Physics 9702 Chapter Notes

What this chapter covers

Capacitance - Capacitors and capacitanceCapacitance - Energy stored in a capacitorCapacitance - Discharging a capacitor
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1. Defining Capacitance

A capacitor is an electrical component that stores energy in an electric field. It typically consists of two conductive plates separated by an insulating material called a dielectric. When connected to a power source like a battery, positive charge builds up on one plate and an equal amount of negative charge builds up on the other. Capacitance (C) is a measure of a capacitor's ability to store charge. It is defined as the amount of charge (Q) stored on each plate for every one volt of potential difference (V) across the plates. The relationship is given by the formula Q = CV. The unit of capacitance is the farad (F). One farad is a very large unit, so capacitance is often measured in microfarads (1 μF = 10⁻⁶ F), nanofarads (1 nF = 10⁻⁹ F), or picofarads (1 pF = 10⁻¹² F).

C = Q / V

Key term

Capacitance: The ratio of the electric charge stored on a conductor to the potential difference across it.

Common pitfall

Confusing the symbol 'C' for capacitance with the unit 'C' for coulombs. Always check the context of the equation or statement.

Fun fact

A 1 Farad capacitor would be enormous if built with traditional parallel plates! The Earth's capacitance is only about 710 μF. Modern supercapacitors can pack several thousand farads into a small cylinder by using materials with huge surface areas, like activated carbon.

Worked example 13 marks

A capacitor is labelled 470 μF. Calculate the charge it stores when the potential difference across it is 12 V.

  1. 1

    Step 1: State the formula relating charge, capacitance and voltage. Q = CV.

  2. 2

    Step 2: Identify the given values. C = 470 μF = 470 × 10⁻⁶ F. V = 12 V.

  3. 3

    Step 3: Substitute the values into the formula. Q = (470 × 10⁻⁶ F) × (12 V).

  4. 4

    Step 4: Calculate the result. Q = 0.00564 C.

  5. 5

    Step 5: Express the answer in appropriate units, for example, millicoulombs (mC). Q = 5.64 mC.

Worked example 23 marks

A capacitor stores 2.5 mC of charge when connected to a 50 V supply. What is its capacitance?

  1. 1

    Step 1: State the formula for capacitance. C = Q / V.

  2. 2

    Step 2: Identify the given values. Q = 2.5 mC = 2.5 × 10⁻³ C. V = 50 V.

  3. 3

    Step 3: Substitute the values into the formula. C = (2.5 × 10⁻³ C) / (50 V).

  4. 4

    Step 4: Calculate the result. C = 0.00005 F.

  5. 5

    Step 5: Express the answer in a more convenient unit, like microfarads (μF). C = 50 × 10⁻⁶ F = 50 μF.

Recap

  • A capacitor stores electrical energy by separating charge on two conductive plates.
  • Capacitance (C) is the charge stored (Q) per unit of potential difference (V).
  • The defining formula is C = Q / V, often rearranged as Q = CV.
  • The unit of capacitance is the Farad (F), where 1 F = 1 C V⁻¹.
  • Commonly used prefixes are micro (μ), nano (n), and pico (p).

Quick check

  1. What is the potential difference across a 2200 μF capacitor that is storing 33 mC of charge?2 marks
  2. Define the farad.1 mark

2. Energy Stored in a Capacitor

To charge a capacitor, work must be done to move electrons from the positive plate to the negative plate against the growing potential difference. This work is stored as electric potential energy in the electric field between the plates. We can find the energy stored by considering the graph of potential difference (V) against charge (Q). As charge is added, the potential difference increases proportionally (V = Q/C), so the graph is a straight line through the origin. The total energy stored (W) is equal to the area under this graph. The area of the triangle is ½ × base × height, which gives W = ½ QV. By substituting Q = CV or V = Q/C, we can derive two other useful forms of the energy equation: W = ½ CV² and W = ½ Q²/C. You should be able to use and derive all three forms.

W = 1/2 QV

W = 1/2 CV²

W = 1/2 Q²/C

Key term

Electric Potential Energy (in a capacitor): The energy stored in a capacitor due to the work done to move charge against the electric field between its plates.

Examiner insight

Examiners often ask for the derivation of the energy formula from a V-Q graph. Be prepared to state that the energy stored is the area under the graph and use the formula for the area of a triangle.

Fun fact

The flash in a disposable camera is powered by a capacitor. A small battery charges the capacitor slowly, and then the capacitor discharges all its stored energy very quickly through the flashbulb, creating a bright burst of light.

Worked example 13 marks

A 1000 μF capacitor is charged by a 9.0 V battery. Calculate the energy stored in the capacitor.

  1. 1

    Step 1: Choose the appropriate energy formula. We are given C and V, so W = ½ CV² is the best choice.

  2. 2

    Step 2: Identify the values. C = 1000 μF = 1000 × 10⁻⁶ F. V = 9.0 V.

  3. 3

    Step 3: Substitute the values into the formula. W = ½ × (1000 × 10⁻⁶ F) × (9.0 V)².

  4. 4

    Step 4: Calculate the result. W = ½ × (1000 × 10⁻⁶) × 81 = 0.0405 J.

  5. 5

    Step 5: Express the answer with appropriate significant figures. W = 0.041 J (or 41 mJ).

Worked example 24 marks

A capacitor stores 5.0 mJ of energy. The charge on its plates is 100 μC. Calculate the capacitance.

  1. 1

    Step 1: Choose the appropriate energy formula. We have W and Q, and want to find C. The formula W = ½ Q²/C is most direct.

  2. 2

    Step 2: Rearrange the formula to make C the subject. C = Q² / (2W).

  3. 3

    Step 3: Identify the values in SI units. W = 5.0 mJ = 5.0 × 10⁻³ J. Q = 100 μC = 100 × 10⁻⁶ C.

  4. 4

    Step 4: Substitute the values. C = (100 × 10⁻⁶ C)² / (2 × 5.0 × 10⁻³ J).

  5. 5

    Step 5: Calculate the result. C = (1 × 10⁻⁸) / (1 × 10⁻²) = 1 × 10⁻⁶ F.

  6. 6

    Step 6: Express the answer in a convenient unit. C = 1.0 μF.

Recap

  • Energy is stored in a capacitor because work is done to separate charges.
  • The energy stored (W) is equal to the area under a potential difference-charge (V-Q) graph.
  • The three key formulas for stored energy are W = ½ QV, W = ½ CV², and W = ½ Q²/C.
  • Choose the energy formula based on the quantities given in the problem.
  • Energy stored is proportional to the square of the voltage or the square of the charge.

Quick check

  1. A capacitor stores 2.5 mJ of energy when the p.d. across it is 100 V. What is its capacitance?2 marks
  2. If the voltage across a capacitor is doubled, by what factor does the stored energy increase?1 mark

3. Capacitors in Parallel

When capacitors are connected in parallel, the 'top' plates of all capacitors are connected together, and the 'bottom' plates are all connected together. This means the potential difference (V) across each capacitor is the same as the supply voltage. The total charge stored (Q_total) is the sum of the charges stored on each individual capacitor (Q_total = Q₁ + Q₂ + ...). Since Q = CV, we can write C_total * V = C₁V + C₂V + ... . Because V is the same for all terms, we can divide it out, leaving the simple rule for combining capacitors in parallel: C_total = C₁ + C₂ + ... . The total capacitance is simply the sum of the individual capacitances. This means adding a capacitor in parallel always increases the total capacitance of the circuit.

C_total = C₁ + C₂ + C₃ + ...

Key term

Parallel Combination: A circuit arrangement where components are connected across the same two points, resulting in the same potential difference across each component.

Common pitfall

Mixing up the rules for capacitors and resistors. For capacitors in parallel, you add the capacitances, which is the same rule as for resistors in series.

Worked example 14 marks

A 220 μF capacitor and a 470 μF capacitor are connected in parallel to a 6.0 V supply. Calculate:(a) the total capacitance,(b) the total charge stored.

  1. 1

    Part (a): Total Capacitance

  2. 2

    Step 1: State the rule for capacitors in parallel. C_total = C₁ + C₂.

  3. 3

    Step 2: Substitute the values. C_total = 220 μF + 470 μF.

  4. 4

    Step 3: Calculate the result. C_total = 690 μF.

  5. 5

    Part (b): Total Charge Stored

  6. 6

    Step 4: Use the formula Q_total = C_total × V.

  7. 7

    Step 5: Substitute the values, ensuring C is in Farads. C_total = 690 μF = 690 × 10⁻⁶ F. V = 6.0 V.

  8. 8

    Step 6: Calculate the charge. Q_total = (690 × 10⁻⁶ F) × (6.0 V) = 4.14 × 10⁻³ C.

  9. 9

    Step 7: Express in a suitable unit. Q_total = 4.1 mC.

Recap

  • In a parallel combination, the potential difference across each capacitor is identical.
  • The total charge stored is the sum of the charges on the individual capacitors.
  • The total capacitance is found by adding the individual capacitances: C_total = C₁ + C₂ + ...
  • Adding a capacitor in parallel always increases the total capacitance.

Quick check

  1. What is the total capacitance of a 10 nF, a 22 nF and a 47 nF capacitor connected in parallel?1 mark
  2. If three identical capacitors are connected in parallel, what is the total capacitance compared to a single one?1 mark

4. Capacitors in Series

When capacitors are connected in series, they are placed one after another in a single line. In this arrangement, the key principles are: 1) The total potential difference from the supply (V_total) is shared between the capacitors (V_total = V₁ + V₂ + ...). 2) The charge (Q) stored on each capacitor is the same. This is because the charging process effectively moves charge from the outer plate of the first capacitor to the outer plate of the last one, inducing equal and opposite charges on the inner plates. Since V = Q/C, we can substitute this into the voltage equation: Q/C_total = Q/C₁ + Q/C₂ + ... . The charge Q is the same for all terms, so we can cancel it, giving the rule for capacitors in series: 1/C_total = 1/C₁ + 1/C₂ + ... . Remember to take the reciprocal of your final answer to find C_total. Adding a capacitor in series always decreases the total capacitance.

1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + ...

Key term

Series Combination: A circuit arrangement where components are connected end-to-end, so the same magnitude of charge is stored on each component.

Examiner insight

Derivations for series capacitance must clearly state that charge is the same on each capacitor and the total p.d. is the sum of individual p.d.s. Marks are awarded for this physical reasoning, not just for writing down the formula.

Common pitfall

After calculating the value of 1/C_total, students often forget to take the reciprocal to find C_total. Always remember this final inversion step.

Worked example 15 marks

A 100 μF capacitor and a 200 μF capacitor are connected in series to a 12 V supply. Calculate:(a) the total capacitance,(b) the charge on each capacitor.

  1. 1

    Part (a): Total Capacitance

  2. 2

    Step 1: State the rule for capacitors in series. 1/C_total = 1/C₁ + 1/C₂.

  3. 3

    Step 2: Substitute the values. 1/C_total = 1/100 + 1/200. (We can work in μF for now).

  4. 4

    Step 3: Find a common denominator. 1/C_total = 2/200 + 1/200 = 3/200.

  5. 5

    Step 4: Invert the fraction to find C_total. C_total = 200 / 3 = 66.7 μF.

  6. 6

    Part (b): Charge on each capacitor

  7. 7

    Step 5: First find the total charge using Q_total = C_total × V.

  8. 8

    Step 6: Substitute values. C_total = 66.7 × 10⁻⁶ F. V = 12 V. Q_total = (66.7 × 10⁻⁶) × 12 = 8.0 × 10⁻⁴ C.

  9. 9

    Step 7: State the charge on each capacitor. In a series circuit, the charge is the same on each component. So, Q₁ = Q₂ = Q_total = 8.0 × 10⁻⁴ C (or 800 μC).

Recap

  • In a series combination, the charge stored on each capacitor is identical.
  • The total potential difference is the sum of the p.d.s across individual capacitors.
  • The total capacitance is found using the reciprocal formula: 1/C_total = 1/C₁ + 1/C₂ + ...
  • Adding a capacitor in series always decreases the total capacitance.
  • Remember to take the reciprocal at the end of the calculation to find C_total.

Quick check

  1. What is the total capacitance of two 100 μF capacitors connected in series?2 marks
  2. Why does adding a capacitor in series decrease the total capacitance?2 marks

5. Capacitor Discharge and the Time Constant

When a charged capacitor is connected across a resistor, it discharges. Electrons flow from the negative plate, through the resistor, to the positive plate until the plates are neutral. This flow of charge is a current. At the start (t=0), the potential difference (V) and charge (Q) are at their maximum, so the initial current (I = V/R) is also at its maximum. As the capacitor discharges, Q decreases, which causes V to decrease (since V=Q/C). As V decreases, the current I also decreases (since I=V/R). The result is that the charge, voltage, and current all fall over time with a distinctive curve called an exponential decay. The rate of this decay is determined by the 'time constant' of the circuit, given the symbol τ (tau). The time constant is the product of the resistance and capacitance: τ = CR. It represents the time taken for the charge, voltage, or current to fall to 1/e (approximately 37%) of its initial value. A large time constant (from a large R or C) means a slow discharge.

τ = CR

Key term

Time Constant (τ): The product of capacitance and resistance (CR) in a circuit, representing the time taken for the charge, potential difference, or current to fall to approximately 37% of its initial value during discharge.

Examiner insight

Be able to sketch and label the exponential decay graphs for charge, p.d., and current against time. You should also be able to mark the position of the time constant on the time axis and show that the value has dropped to 37% of the initial value.

Fun fact

Capacitor-resistor circuits are the heart of simple timing devices. They control the flashing rate of LEDs, the timing of intermittent windscreen wipers, and the 'pings' in old-school sonar systems.

Worked example 14 marks

A 2200 μF capacitor is discharged through a 5.0 kΩ resistor.(a) Calculate the time constant of the circuit.(b) If the initial p.d. was 12 V, what is the p.d. after one time constant?

  1. 1

    Part (a): Calculate the time constant

  2. 2

    Step 1: State the formula for the time constant. τ = CR.

  3. 3

    Step 2: Identify the values in SI units. C = 2200 μF = 2200 × 10⁻⁶ F. R = 5.0 kΩ = 5.0 × 10³ Ω.

  4. 4

    Step 3: Substitute the values. τ = (2200 × 10⁻⁶ F) × (5.0 × 10³ Ω).

  5. 5

    Step 4: Calculate the result. τ = 11 s.

  6. 6

    Part (b): Find the p.d. after one time constant

  7. 7

    Step 5: Recall the definition of the time constant. It's the time for the quantity to fall to 1/e (or ~37%) of its initial value.

  8. 8

    Step 6: Calculate the p.d. V = V₀ / e = 12 V / e.

  9. 9

    Step 7: V ≈ 12 V × 0.37 = 4.44 V. The p.d. will be approximately 4.4 V.

Recap

  • When a capacitor discharges through a resistor, Q, V, and I all decrease exponentially.
  • The rate of discharge is determined by the time constant, τ = CR.
  • A large time constant (large C or R) means a slow discharge, while a small time constant means a fast discharge.
  • The time constant is the time for the value (Q, V, or I) to fall to 1/e (about 37%) of its initial value.
  • After 5 time constants (5τ), a capacitor is considered to be almost fully discharged (>99%).

Quick check

  1. Calculate the time constant of a circuit with a 100 μF capacitor and a 47 kΩ resistor.2 marks
  2. A capacitor discharges through a resistor. After one time constant, what percentage of the initial charge remains?1 mark

6. The Mathematics of Exponential Decay

The exponential decay of a discharging capacitor can be described precisely with a single mathematical formula. For any quantity 'x' (which can be charge Q, potential difference V, or current I) that is decaying, its value at time 't' is given by: x = x₀e^(-t/RC). In this equation, x₀ is the initial value of the quantity at t=0, 'e' is the base of the natural logarithm (approximately 2.718), and RC is the time constant (τ) of the circuit. This equation allows you to calculate the charge, voltage, or current at any moment during the discharge. For example, the potential difference is V = V₀e^(-t/RC). You can also rearrange this equation to find the time it takes for a quantity to fall to a certain level. This requires using natural logarithms (ln). Taking the natural log of both sides of V/V₀ = e^(-t/RC) gives ln(V/V₀) = -t/RC, which can be rearranged to find t: t = -RC ln(V/V₀).

x = x₀e^(-t/RC)

Q = Q₀e^(-t/RC)

V = V₀e^(-t/RC)

I = I₀e^(-t/RC)

t = -RC ln(x/x₀)

Key term

Exponential Decay: A process where a quantity decreases at a rate proportional to its current value, described mathematically by an exponential function with a negative exponent.

Examiner insight

Questions often require you to rearrange the decay equation using natural logarithms to find the time 't'. Show your rearrangement steps clearly, as marks are awarded for the method as well as the final answer.

Fun fact

The concept of 'half-life' in radioactive decay is mathematically identical to capacitor discharge. The half-life (t½) of a capacitor's charge is the time it takes to fall to 50% and is related to the time constant by t½ = RC ln(2) ≈ 0.693 RC.

Worked example 14 marks

A 470 μF capacitor is charged to 12 V and then discharged through a 20 kΩ resistor. What is the potential difference across the capacitor after 10 seconds?

  1. 1

    Step 1: First, calculate the time constant, τ = RC.

  2. 2

    Step 2: τ = (20 × 10³ Ω) × (470 × 10⁻⁶ F) = 9.4 s.

  3. 3

    Step 3: State the decay equation for voltage: V = V₀e^(-t/RC).

  4. 4

    Step 4: Substitute the known values: V₀ = 12 V, t = 10 s, RC = 9.4 s.

  5. 5

    Step 5: V = 12 × e^(-10 / 9.4) = 12 × e^(-1.064).

  6. 6

    Step 6: Calculate the value. V = 12 × 0.345 = 4.14 V.

  7. 7

    Step 7: The potential difference is 4.1 V (to 2 s.f.).

Worked example 24 marks

For the same circuit as above (C=470 μF, R=20 kΩ, V₀=12 V), how long does it take for the charge on the capacitor to fall to 25% of its initial value?

  1. 1

    Step 1: We want to find the time t when Q = 0.25 Q₀.

  2. 2

    Step 2: Use the decay equation Q = Q₀e^(-t/RC). This gives 0.25 Q₀ = Q₀e^(-t/RC).

  3. 3

    Step 3: Cancel Q₀ to get 0.25 = e^(-t/RC).

  4. 4

    Step 4: Take the natural logarithm (ln) of both sides: ln(0.25) = -t/RC.

  5. 5

    Step 5: Rearrange to solve for t: t = -RC × ln(0.25).

  6. 6

    Step 6: We already know RC = 9.4 s. ln(0.25) = -1.386.

  7. 7

    Step 7: t = -9.4 s × (-1.386) = 13.03 s.

  8. 8

    Step 8: The time taken is 13 s (to 2 s.f.).

Recap

  • The exponential decay of charge, p.d., and current is described by x = x₀e^(-t/RC).
  • x₀ is the initial value at time t=0 and RC is the time constant τ.
  • This single equation works for charge (Q), potential difference (V), and current (I).
  • To find the time 't' for a quantity to fall from x₀ to x, use the relationship t = -RC ln(x/x₀).
  • Ensure all units are consistent (Farads, Ohms, seconds) before calculation.

Quick check

  1. A capacitor circuit has a time constant of 5.0 s. What fraction of the initial p.d. remains after 5.0 s?2 marks
  2. In the equation t = -RC ln(x/x₀), what would be the value of x/x₀ if we are finding the half-life of the decay?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define capacitance and state its SI unit.2 marks
  2. A 50 μF capacitor is connected to a 12 V d.c. supply. Calculate the charge stored on the capacitor and the energy it stores.4 marks
  3. Three capacitors with capacitances 100 μF, 200 μF and 300 μF are available. Calculate the maximum and minimum total capacitance that can be achieved by combining all three.5 marks
  4. A 10 μF capacitor and a 40 μF capacitor are connected in series to a 50 V supply. Calculate (i) the total capacitance, (ii) the charge on the 10 μF capacitor, and (iii) the potential difference across the 40 μF capacitor.6 marks
  5. A student investigates capacitor discharge. They charge a 2200 μF capacitor to 9.0 V and discharge it through a fixed resistor. The student notes that the potential difference falls to 3.3 V after 25 s. Calculate the resistance of the resistor.5 marks
  6. Sketch a graph to show how the current varies with time as a capacitor discharges through a resistor. On your graph, label the initial current I₀ and indicate the time constant τ.3 marks
  7. A capacitor in a timing circuit must hold its voltage above 4.0 V for at least 10 s. It is charged to an initial voltage of 12 V and discharges through a 150 kΩ resistor. What is the minimum capacitance required?5 marks
  8. Explain, by referring to the movement of charge, why the energy stored in a capacitor is given by W = ½QV and not W = QV.4 marks
  9. A circuit consists of a 10 μF capacitor in series with a 20 μF capacitor. This combination is connected in parallel with a 30 μF capacitor. The entire arrangement is connected to a 12 V supply. Calculate the total energy stored in the circuit.6 marks
  10. A capacitor is charged and then discharged through a sensitive ammeter. The initial discharge current is 80 μA. After 20 s, the current has fallen to 15 μA. The resistor in the circuit has a resistance of 250 kΩ. Calculate the capacitance of the capacitor.5 marks

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