CIE AS/A Level Physics 19.1 Capacitors and capacitance Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 19.1 Capacitors and capacitance Study Notes – New Syllabus
CIE AS/A Level Physics 19.1 Capacitors and capacitance Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- define capacitance, as applied to both isolated spherical conductors and to parallel plate capacitors
- recall and use C = Q / V
- derive, using C = Q / V, formulae for the combined capacitance of capacitors in series and in parallel
- use the capacitance formulae for capacitors in series and in parallel
Definition of Capacitance
Capacitance is a measure of how much electric charge a conductor can store per unit potential difference between its plates or surfaces.
General Definition:
Capacitance is defined as the charge stored per unit potential difference.
\( \mathrm{C = \dfrac{Q}{V}} \)
- \( \mathrm{C} \) = capacitance (F, farad)
- \( \mathrm{Q} \) = charge stored (C)
- \( \mathrm{V} \) = potential difference (V)
Capacitance depends only on the geometry of the conductor(s) and the properties of the medium between them (for parallel plates).
Capacitance of an Isolated Spherical Conductor
For a single isolated conducting sphere of radius \( \mathrm{R} \):
\( \mathrm{C = 4\pi\epsilon_0 R} \)
- Bigger radius → larger capacitance.
- Capacitance increases if placed in a medium with higher permittivity.
- This formula comes from using \( \mathrm{V = \dfrac{Q}{4\pi\epsilon_0 R}} \) and \( \mathrm{C = Q/V} \).
Capacitance of a Parallel Plate Capacitor
For two parallel plates of area \( \mathrm{A} \) separated by distance \( \mathrm{d} \) and filled with air/vacuum:
\( \mathrm{C = \dfrac{\epsilon_0 A}{d}} \)
- Increasing plate area → more charge can be stored → larger capacitance.
- Increasing separation → weaker field → smaller capacitance.
- If a dielectric is inserted, \( \mathrm{\epsilon_0} \) becomes \( \mathrm{\epsilon = \epsilon_r \epsilon_0} \), increasing capacitance.
Example
A spherical conductor has radius \( \mathrm{0.20\ m} \). Calculate its capacitance.
▶️ Answer / Explanation
Use:
\( \mathrm{C = 4\pi\epsilon_0 R} \)
\( \mathrm{C = 4\pi(8.85\times10^{-12})(0.20)} \)
\( \mathrm{C = 2.22\times10^{-11}\ F} \)
Capacitance ≈ \( \mathrm{2.2\times10^{-11}\ F} \)
Example
Two parallel plates each have area \( \mathrm{0.04\ m^2} \) and are separated by \( \mathrm{2.0\ mm} \). Calculate their capacitance in air.
▶️ Answer / Explanation
Convert distance:
\( \mathrm{d = 2.0\ mm = 2.0\times10^{-3}\ m} \)
Use:
\( \mathrm{C = \dfrac{\epsilon_0 A}{d}} \)
\( \mathrm{C = \dfrac{(8.85\times10^{-12})(0.04)}{2.0\times10^{-3}}} \)
\( \mathrm{C = 1.77\times10^{-10}\ F} \)
Capacitance = \( \mathrm{1.8\times10^{-10}\ F} \)
Example
A parallel plate capacitor has area \( \mathrm{A = 0.06\ m^2} \), separation \( \mathrm{d = 1.5\ mm} \). A dielectric with \( \mathrm{\epsilon_r = 4} \) completely fills the space. Calculate the new capacitance.
▶️ Answer / Explanation
Effective permittivity:
\( \mathrm{\epsilon = \epsilon_r \epsilon_0 = 4(8.85\times10^{-12}) = 3.54\times10^{-11}} \)
Use:
\( \mathrm{C = \dfrac{\epsilon A}{d}} \)
\( \mathrm{C = \dfrac{(3.54\times10^{-11})(0.06)}{1.5\times10^{-3}}} \)
\( \mathrm{C = 1.42\times10^{-9}\ F} \)
Capacitance = \( \mathrm{1.4\times10^{-9}\ F} \)
Capacitance Formula: \( \mathrm{C = \dfrac{Q}{V}} \)
The basic definition of capacitance relates the charge stored on a capacitor to the potential difference across it.
Definition:
Capacitance is the charge stored per unit potential difference.
\( \mathrm{C = \dfrac{Q}{V}} \)
- \( \mathrm{C} \) = capacitance (farads, F)
- \( \mathrm{Q} \) = charge stored (coulombs, C)
- \( \mathrm{V} \) = potential difference (volts, V)
Interpretation:
- A capacitor with large capacitance stores more charge for each volt applied.
- If the voltage increases while capacitance is constant, the stored charge increases proportionally.
- Capacitance is independent of charge or voltage – it is a property of the capacitor.
Example
A capacitor stores \( \mathrm{3.0\times10^{-6}\ C} \) of charge when the potential difference across it is 6.0 V. Calculate its capacitance.
▶️ Answer / Explanation
Use the formula:
\( \mathrm{C = \dfrac{Q}{V}} \)
\( \mathrm{C = \dfrac{3.0\times10^{-6}}{6.0}} \)
\( \mathrm{C = 5.0\times10^{-7}\ F} \)
Capacitance = \( \mathrm{5.0\times10^{-7}\ F} \)
Example
A 220 μF capacitor is connected across a 12 V supply. How much charge does it store?
▶️ Answer / Explanation
Convert capacitance:
\( \mathrm{220\ \mu F = 220\times10^{-6}\ F} \)
Use:
\( \mathrm{Q = CV} \)
\( \mathrm{Q = (220\times10^{-6})(12)} \)
\( \mathrm{Q = 2.64\times10^{-3}\ C} \)
Charge stored = \( \mathrm{2.64\times10^{-3}\ C} \)
Example
A capacitor stores a charge of \( \mathrm{1.5\times10^{-3}\ C} \) when the voltage across it is 50 V. If the voltage is increased to 150 V with no change in capacitance, how much total charge will it store?
▶️ Answer / Explanation
Step 1: Find capacitance
\( \mathrm{C = \dfrac{Q}{V} = \dfrac{1.5\times10^{-3}}{50}} \)
\( \mathrm{C = 3.0\times10^{-5}\ F} \)
Step 2: Find charge at 150 V
\( \mathrm{Q = CV = (3.0\times10^{-5})(150)} \)
\( \mathrm{Q = 4.5\times10^{-3}\ C} \)
Total charge stored = \( \mathrm{4.5\times10^{-3}\ C} \)
Deriving Formulae for Combined Capacitance in Series and Parallel
Using the fundamental definition of capacitance:
\( \mathrm{C = \dfrac{Q}{V}} \)
we can derive the expressions for the equivalent capacitance of capacitors connected in series and in parallel.
Capacitors in Parallel
Consider two capacitors \( \mathrm{C_1} \) and \( \mathrm{C_2} \) connected in parallel across a supply of voltage \( \mathrm{V} \).
- The potential difference across each capacitor is the same: \( \mathrm{V_1 = V_2 = V} \)
- Total charge stored: \( \mathrm{Q = Q_1 + Q_2} \)
- Using \( \mathrm{Q = CV} \): \( \mathrm{Q_1 = C_1 V} \), \( \mathrm{Q_2 = C_2 V} \)
Derivation:
\( \mathrm{Q = Q_1 + Q_2 = C_1 V + C_2 V = (C_1 + C_2)V} \)
But the total charge also equals:
\( \mathrm{Q = C_{\text{eq}} V} \)
Therefore:
\( \mathrm{C_{\text{eq}} = C_1 + C_2 + \cdots} \)
Capacitors in parallel add directly.
Capacitors in Series
Now consider two capacitors \( \mathrm{C_1} \) and \( \mathrm{C_2} \) connected in series.
- They each carry the same charge: \( \mathrm{Q_1 = Q_2 = Q} \)
- Total potential difference across the combination: \( \mathrm{V = V_1 + V_2} \)
- Using \( \mathrm{V = Q/C} \): \( \mathrm{V_1 = Q/C_1} \), \( \mathrm{V_2 = Q/C_2} \)
Derivation:
\( \mathrm{V = V_1 + V_2 = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}} \)
But:
\( \mathrm{V = \dfrac{Q}{C_{\text{eq}}}} \)
So:
\( \mathrm{\dfrac{Q}{C_{\text{eq}}} = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}} \)
Divide both sides by \( \mathrm{Q} \):
\( \mathrm{\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots} \)
Reciprocal addition is used for capacitors in series.
Example
Two capacitors, \( \mathrm{C_1 = 4\ \mu F} \) and \( \mathrm{C_2 = 6\ \mu F} \), are connected in parallel. Find the equivalent capacitance.
▶️ Answer / Explanation
Use:
\( \mathrm{C_{\text{eq}} = C_1 + C_2} \)
\( \mathrm{C_{\text{eq}} = 4 + 6 = 10\ \mu F} \)
Equivalent capacitance = \( \mathrm{10\ \mu F} \)
Example
Two capacitors, \( \mathrm{C_1 = 12\ \mu F} \) and \( \mathrm{C_2 = 8\ \mu F} \), are connected in series. Find the equivalent capacitance.
▶️ Answer / Explanation
Use:
\( \mathrm{\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{12} + \dfrac{1}{8}} \)
\( \mathrm{\dfrac{1}{C_{\text{eq}}} = 0.0833 + 0.125 = 0.2083} \)
Invert:
\( \mathrm{C_{\text{eq}} = 4.8\ \mu F} \)
Equivalent capacitance = \( \mathrm{4.8\ \mu F} \)
Example
Three capacitors are connected as follows: Two capacitors, \( \mathrm{C_1 = 5\ \mu F} \) and \( \mathrm{C_2 = 10\ \mu F} \), are in series. This combination is connected in parallel with a capacitor \( \mathrm{C_3 = 8\ \mu F} \). Find the total capacitance.
▶️ Answer / Explanation
Step 1: Combine series capacitors
\( \mathrm{\dfrac{1}{C_s} = \dfrac{1}{5} + \dfrac{1}{10} = 0.2 + 0.1 = 0.3} \)
\( \mathrm{C_s = \dfrac{1}{0.3} = 3.33\ \mu F} \)
Step 2: Add parallel capacitor
\( \mathrm{C_{\text{eq}} = C_s + C_3 = 3.33 + 8.0 = 11.33\ \mu F} \)
Total capacitance = \( \mathrm{11.33\ \mu F} \)
Using the Capacitance Formulae for Series and Parallel Connections
Once the formulae for combining capacitors are known, they can be applied to calculate the equivalent capacitance of any arrangement of capacitors.
Key Formulae:
Parallel: \( \mathrm{C_{\text{eq}} = C_1 + C_2 + C_3 + \cdots} \)
Series: \( \mathrm{\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \cdots} \)
- Parallel → capacitances add directly.
- Series → reciprocals add.
- Mixed arrangements → simplify step-by-step.
Example
Two capacitors, \( \mathrm{C_1 = 3\ \mu F} \) and \( \mathrm{C_2 = 7\ \mu F} \), are connected in parallel. Calculate the total capacitance.
▶️ Answer / Explanation
Parallel → add directly:
\( \mathrm{C_{\text{eq}} = C_1 + C_2 = 3 + 7 = 10\ \mu F} \)
Total capacitance = \( \mathrm{10\ \mu F} \)
Example
Two capacitors, \( \mathrm{C_1 = 12\ \mu F} \) and \( \mathrm{C_2 = 4\ \mu F} \), are connected in series. Calculate the equivalent capacitance.
▶️ Answer / Explanation
Series → reciprocal formula:
\( \mathrm{\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{12} + \dfrac{1}{4}} \)
\( \mathrm{\dfrac{1}{C_{\text{eq}}} = 0.0833 + 0.25 = 0.3333} \)
Invert:
\( \mathrm{C_{\text{eq}} = 3.0\ \mu F} \)
Equivalent capacitance = \( \mathrm{3.0\ \mu F} \)
Example
Three capacitors are arranged as follows:
• \( \mathrm{C_1 = 8\ \mu F} \) and \( \mathrm{C_2 = 12\ \mu F} \) are connected in series.
• Their series combination is then connected in parallel with a capacitor \( \mathrm{C_3 = 5\ \mu F} \).
Find the total capacitance of the arrangement.
▶️ Answer / Explanation
Step 1: Combine series capacitors \( \mathrm{C_1} \) and \( \mathrm{C_2} \)
\( \mathrm{\dfrac{1}{C_s} = \dfrac{1}{8} + \dfrac{1}{12}} \)
LCM of 8 and 12 = 24 → \( \mathrm{\dfrac{1}{C_s} = \dfrac{3}{24} + \dfrac{2}{24} = \dfrac{5}{24}} \)
\( \mathrm{C_s = \dfrac{24}{5} = 4.8\ \mu F} \)
Step 2: Add parallel capacitor
\( \mathrm{C_{\text{eq}} = C_s + C_3 = 4.8 + 5.0 = 9.8\ \mu F} \)
Total capacitance = \( \mathrm{9.8\ \mu F} \)