Edexcel A Level (IAL) Physics-4.20 Capacitance- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -4.20 Capacitance- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -4.20 Capacitance- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
Capacitance and the Equation \( C = \dfrac{Q}{V} \)
Capacitance is a measure of how much electric charge a device can store for a given potential difference. It is a key concept in the study of capacitors and electrical energy storage.
Definition of Capacitance
Capacitance is defined as the charge stored per unit potential difference.
\( C = \dfrac{Q}{V} \)
- \( C \) = capacitance (farad, F)
- \( Q \) = charge stored (coulomb, C)
- \( V \) = potential difference (volt, V)
Meaning of the Equation
- A larger capacitance means more charge can be stored for the same voltage.
- If the potential difference increases, the stored charge increases proportionally (for a fixed capacitor).
- Capacitance depends on the physical construction of the capacitor, not on \( Q \) or \( V \) individually.
Unit of Capacitance
The SI unit of capacitance is the farad (F).
\( 1\ \mathrm{F} = 1\ \mathrm{C\,V^{-1}} \)
Practical note:
- The farad is a very large unit.
- Common capacitors are measured in microfarads (µF), nanofarads (nF), or picofarads (pF).
Rearranging the Capacitance Equation
From \( C = \dfrac{Q}{V} \):
\( Q = CV \)
\( V = \dfrac{Q}{C} \)
These forms are used depending on which quantities are known.
Important Points to Remember
- Capacitance is constant for a given capacitor.
- Doubling the voltage doubles the stored charge.
- Capacitance does not depend on the amount of charge stored.
- Capacitors store energy in the electric field between their plates.
Example (Easy)
A capacitor stores a charge of \( 4.0\times10^{-6}\ \mathrm{C} \) when connected to a \( 2.0\ \mathrm{V} \) supply. Calculate its capacitance.
▶️ Answer / Explanation
\( C = \dfrac{Q}{V} = \dfrac{4.0\times10^{-6}}{2.0} = 2.0\times10^{-6}\ \mathrm{F} \)
Capacitance = \( 2.0\ \mathrm{\mu F} \)
Example (Medium)
A capacitor of capacitance \( 470\ \mathrm{\mu F} \) is connected across a \( 9.0\ \mathrm{V} \) battery. Find the charge stored.
▶️ Answer / Explanation
Use \( Q = CV \):
\( Q = 470\times10^{-6} \times 9.0 = 4.23\times10^{-3}\ \mathrm{C} \)
Example (Hard)
A capacitor stores \( 6.0\times10^{-5}\ \mathrm{C} \) of charge when fully charged. If its capacitance is \( 12\ \mathrm{\mu F} \), calculate the potential difference across it.
▶️ Answer / Explanation
Use \( V = \dfrac{Q}{C} \):
\( V = \dfrac{6.0\times10^{-5}}{12\times10^{-6}} = 5.0\ \mathrm{V} \)