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

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:

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}{\varepsilon0\varepsilonr}

\]

3. Potential difference \(V=Ed=\displaystyle\frac{Qd}{\varepsilon0\varepsilonrA}\).
4. Re‑arranging \(C=Q/V\) gives the ideal parallel‑plate expression

\[

\boxed{C=\varepsilon0\varepsilonr\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 \(\varepsilonr\) (often written \(C=\kappa C0\) 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

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.