use the capacitance formulae for capacitors in series and in parallel

Cambridge AS & A Level Physics (9702) – Capacitors and Capacitance

1. Syllabus links & assessment objectives

Syllabus item	Notes covered	Assessment objectives
19.1 – Definition of a capacitor and capacitance	Section 2	AO1
19.2 – Parallel‑plate capacitor formula, dielectrics	Section 3	AO1, AO2
19.3 – Energy stored, RC‑time constant, series & parallel combinations	Sections 4‑6	AO1, AO2, AO3
Practical skills (Paper 3 & 5)	Section 7	AO3

2. What is a capacitor?

• Two conductors (plates) separated by an insulating material (dielectric).
• When a potential difference \(V\) is applied, equal and opposite charges \(+Q\) and \(-Q\) accumulate on the plates.
• The ability to store charge is quantified by the capacitance \(C\):

\[ C=\frac{Q}{V} \]

Units: farad (F), where \(1\;\text{F}=1\;\text{C V}^{-1}\).

Typical ranges in the syllabus:

Device	Typical capacitance
Electrolytic capacitor	10 µF – 1000 µF
Film / ceramic capacitor	0.1 pF – 10 µF

3. Parallel‑plate capacitor – derivation of the ideal formula

1. Surface charge density \(\displaystyle\sigma=\frac{Q}{A}\).
2. Electric field between the plates (ignoring edge effects):
\[ E=\frac{\sigma}{\varepsilon_0\varepsilon_r} \]
3. Potential difference \(V=Ed=\displaystyle\frac{Qd}{\varepsilon_0\varepsilon_rA}\).
4. Re‑arranging \(C=Q/V\) gives the ideal parallel‑plate expression

\[ \boxed{C=\varepsilon_0\varepsilon_r\frac{A}{d}} \]

• \(A\) – plate area (m²)
• \(d\) – separation (m)
• \(\varepsilon_0=8.85\times10^{-12}\;\text{F m}^{-1}\) (vacuum permittivity)
• \(\varepsilon_r\) – relative permittivity (dielectric constant) of the material between the plates.

Important note (edge effects): The formula assumes \(A\gg d\) so that fringe fields are negligible. When this condition is not met the calculated capacitance is an underestimate.

Consequences

• Increasing plate area \(\uparrow A\) → \(\uparrow C\).
• Increasing separation \(\uparrow d\) → \(\downarrow C\).
• Inserting a dielectric multiplies the capacitance by \(\varepsilon_r\) (often written \(C=\kappa C_0\) with \(\kappa=\varepsilon_r\)).

4. Energy stored in a capacitor

The work done in moving charge onto the plates is stored as electrostatic potential energy:

\[ W=\int_{0}^{Q}V\,\mathrm{d}Q =\int_{0}^{Q}\frac{Q'}{C}\,\mathrm{d}Q' =\frac{1}{2}\frac{Q^{2}}{C} =\frac{1}{2}CV^{2} =\frac{1}{2}QV \]

Example (AO1) – A 10 µF capacitor charged to 12 V:

\[ W=\tfrac12CV^{2} =\tfrac12(10\times10^{-6})(12^{2}) =7.2\times10^{-4}\;\text{J}=0.72\;\text{mJ} \]

5. Discharging a capacitor – the RC time constant

When a charged capacitor is connected across a resistor \(R\), the voltage and charge decay exponentially:

\[ V(t)=V_{0}e^{-t/RC},\qquad Q(t)=Q_{0}e^{-t/RC} \]

The product \(\tau =RC\) is the time constant. After one \(\tau\) the voltage (or charge) has fallen to \(\frac{1}{e}\approx0.368\) of its initial value.

Example (AO2) – 4.7 µF capacitor discharging through 2 kΩ:

\[ \tau =RC = (2.0\times10^{3})(4.7\times10^{-6}) = 9.4\times10^{-3}\;\text{s}=9.4\;\text{ms} \] \[ V(20\text{ ms}) = V_{0}e^{-20/9.4}=V_{0}e^{-2.13}=0.12\,V_{0} \]

6. Combining capacitors

6.1 Capacitors in parallel

• All corresponding plates are connected together, so each capacitor experiences the same voltage \(V\).
• The total charge is the algebraic sum of the individual charges.

\[ \boxed{C_{\text{eq}}=\displaystyle\sum_{i=1}^{n}C_{i}} \]

6.2 Capacitors in series

• The same charge \(Q\) passes through each capacitor; the individual voltages add to give the total voltage.
• Working with reciprocals makes the algebra simple.

\[ \boxed{\displaystyle\frac{1}{C_{\text{eq}}}= \displaystyle\sum_{i=1}^{n}\frac{1}{C_{i}}} \]

Derivation (two capacitors, AO2):

\[ V=V_{1}+V_{2}=Q\!\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right) \quad\Longrightarrow\quad C_{\text{eq}}=\frac{Q}{V} \;\Rightarrow\; \frac{1}{C_{\text{eq}}}= \frac{1}{C_{1}}+\frac{1}{C_{2}} \]

6.3 Mixed series‑parallel networks (AO2)

Work from the innermost combination outwards, applying the appropriate rule at each step.

Worked example – \(C_{1}=2\;\mu\text{F},\;C_{2}=3\;\mu\text{F},\;C_{3}=6\;\mu\text{F}\).

1. All three in parallel \[ C_{\text{eq}}=2+3+6=11\;\mu\text{F} \]
2. \(C_{1}\) and \(C_{2}\) in series, then in parallel with \(C_{3}\)

   Series pair:

   \[ \frac{1}{C_{12}}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} \;\Longrightarrow\; C_{12}= \frac{6}{5}=1.2\;\mu\text{F} \]

   Parallel with \(C_{3}\):

   \[ C_{\text{eq}}=C_{12}+C_{3}=1.2+6=7.2\;\mu\text{F} \]

7. Practical measurement of capacitance (Paper 3 & 5 – AO3)

• RC‑timing method: Connect the unknown capacitor in series with a known resistor, charge through a switch, then measure the time for the voltage to fall to \(0.368\,V_{0}\). The time equals \(\tau=RC\); hence \(C=\tau/R\).
• Bridge method (Wheatstone‑type capacitance bridge): Balance a known reference capacitor against the unknown using a variable resistor and a null detector.
• Sources of error:
  • Parasitic resistance of leads (affects \(\tau\)).
  • Leakage through the dielectric, especially for electrolytic types.
  • Temperature dependence of \(\varepsilon_r\).
• Safety note: Never discharge a high‑voltage capacitor directly with a finger; use a resistor (≥ 10 kΩ) to bring the voltage safely to zero.

Design task (AO3)

“Design an experiment to determine the dielectric constant of a sheet of mica placed between the plates of a parallel‑plate capacitor. State the equipment, procedure, how you would calculate \(\varepsilon_r\), and discuss at least two possible sources of systematic error.”

8. Summary of key formulae

Configuration	Formula	Key points
Parallel‑plate (ideal)	\(C=\varepsilon_0\varepsilon_r\dfrac{A}{d}\)	\(\uparrow A\) ↑ \(C\); \(\uparrow d\) ↓ \(C\); dielectric multiplies by \(\varepsilon_r\).
Parallel connection	\(C_{\text{eq}}=\displaystyle\sum C_i\)	Same voltage on each; charges add.
Series connection	\(\displaystyle\frac{1}{C_{\text{eq}}}= \displaystyle\sum\frac{1}{C_i}\)	Same charge on each; voltages add.
Energy stored	\(W=\tfrac12CV^{2}= \tfrac12QV = \tfrac{Q^{2}}{2C}\)	Energy supplied by the source while charging.
RC discharge	\(V(t)=V_{0}e^{-t/RC},\;\;\tau=RC\)	After one \(\tau\) voltage = 0.368 \(V_{0}\).

9. Common mistakes to avoid (AO1)

• Voltage vs. charge handling – In parallel the voltage is the same; in series the charge is the same.
• Using the wrong combination rule – Do not apply the reciprocal rule to a parallel network.
• Neglecting the dielectric constant – Forgetting \(\varepsilon_r\) under‑estimates real capacitance.
• Unit conversion errors – Always convert µF, nF, pF to farads before substituting.
• Edge‑effect assumption – The parallel‑plate formula is only accurate when plate area ≫ separation.

10. Practice questions (with answers)

1. Series charge and voltage (AO2) – Two capacitors, \(C_{1}=4\;\mu\text{F}\) and \(C_{2}=6\;\mu\text{F}\), are connected in series across a 12 V battery. Find the equivalent capacitance, the charge on each capacitor, and the voltage across each.
   Solution

   \[ \frac{1}{C_{\text{eq}}}= \frac{1}{4}+\frac{1}{6}= \frac{5}{12} \;\Longrightarrow\; C_{\text{eq}}= \frac{12}{5}=2.4\;\mu\text{F} \] \[ Q=C_{\text{eq}}V = 2.4\times10^{-6}\times12 = 2.88\times10^{-5}\;\text{C} \] \[ V_{1}= \frac{Q}{C_{1}} = \frac{2.88\times10^{-5}}{4\times10^{-6}} = 7.2\;\text{V} \] \[ V_{2}= \frac{Q}{C_{2}} = \frac{2.88\times10^{-5}}{6\times10^{-6}} = 4.8\;\text{V} \]
2. Mixed network (AO2) – A \(10\;\mu\text{F}\) capacitor is placed in parallel with three \(5\;\mu\text{F}\) capacitors connected in series. Find the total capacitance.
   Solution

   Series of three \(5\;\mu\text{F}\): \[ \frac{1}{C_{s}}= \frac{1}{5}+\frac{1}{5}+\frac{1}{5}= \frac{3}{5} \;\Longrightarrow\; C_{s}= \frac{5}{3}=1.67\;\mu\text{F} \] Parallel with \(10\;\mu\text{F}\): \[ C_{\text{eq}}=10+1.67=11.67\;\mu\text{F} \]
3. Parallel‑plate calculation (AO1) – Plate area \(A=0.02\;\text{m}^{2}\), separation \(d=1.5\;\text{mm}\), air dielectric (\(\varepsilon_{r}\approx1\)). Compute \(C\).
   Solution

   \[ C=\varepsilon_{0}\varepsilon_{r}\frac{A}{d} =(8.85\times10^{-12})(1)\frac{0.02}{1.5\times10^{-3}} =1.18\times10^{-10}\;\text{F}=118\;\text{pF} \]
4. RC discharge (AO2) – A 22 µF capacitor discharges through a 470 kΩ resistor.
   • Calculate the time constant \(\tau\).
   • What is the voltage after 0.5 s if the initial voltage is 15 V?
   Solution

   \[ \tau =RC = (4.7\times10^{5})(22\times10^{-6}) = 10.34\;\text{s} \] \[ V(0.5\text{ s}) = 15\,e^{-0.5/10.34}=15\,e^{-0.048}=14.3\;\text{V} \]
5. Energy comparison (AO1) – Which stores more energy: (a) 5 µF at 20 V or (b) 10 µF at 12 V?
   Solution

   \[ W_{a}= \tfrac12(5\times10^{-6})(20^{2}) = 1.0\times10^{-3}\;\text{J} \] \[ W_{b}= \tfrac12(10\times10^{-6})(12^{2}) = 0.72\times10^{-3}\;\text{J} \]

   Capacitor (a) stores more energy.