Capacitors and capacitance Edexcel International A Level Physics revision
Not started
Learn it
In plain words
A capacitor is a pair of metal plates with a gap between them. Connect a battery and charge piles up, positive on one plate and negative on the other, each held in place by the attraction of the other across the gap. The capacitance tells you how much charge is stored for each volt.
Three things to know
- Capacitance C = Q ÷ V: the charge stored per unit potential difference. Unit: the farad (F), which is one coulomb per volt.
- Real capacitors are measured in microfarads (µF, 10⁻⁶ F), nanofarads (nF, 10⁻⁹ F) or picofarads (pF, 10⁻¹² F).
- Q is the charge on one plate. One plate has +Q and the other has −Q.
Worked example
A 470 µF capacitor is charged to a p.d. of 12 V. Find the charge stored.
- Q = CV.
- = 470 × 10⁻⁶ × 12.
- = 5.6 × 10⁻³ C.
Tips and tricks
- Change microfarads to farads before you calculate.
- A capacitor is charged at a steady current I for a time t: the charge is Q = It.
It lands in your notebook with its questions as flashcards.
Capacitors and capacitance: 5 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
What is capacitance?
C = Q ÷ V.
What is the unit of capacitance?
One farad is one coulomb per volt.
A 100 µF capacitor has a p.d. of 5.0 V across it. How much charge does it store?
Q = CV.
The p.d. across a capacitor is doubled. What happens to the charge it stores?
The capacitance stays the same, and Q = CV.
What is 1 µF in farads?
Micro means 10⁻⁶.
Quiz
5 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
3 questions, 7 marks. Write your answers on paper, then check them.
Capacitors and capacitance
Edexcel International A Level Physics WPH · 7 marks · papermunch.org
Name ______________________________ Date ______________
A capacitor stores a charge of 3.0 × 10⁻⁴ C when the p.d. across it is 6.0 V. Find its capacitance.[2]
Show answerHide answer
5.0 × 10⁻⁵ F (50 µF).
Find the p.d. across a 2.0 µF capacitor that holds a charge of 9.0 µC.[2]
Show answerHide answer
4.5 V.
A capacitor is charged by a steady current of 20 µA for 30 s, and the p.d. across it reaches 3.0 V. Find its capacitance.[3]
Show answerHide answer
2.0 × 10⁻⁴ F (200 µF). The charge is 20 × 10⁻⁶ × 30 = 6.0 × 10⁻⁴ C.
Answers: Capacitors and capacitance
- 1. 5.0 × 10⁻⁵ F (50 µF).
- 2. 4.5 V.
- 3. 2.0 × 10⁻⁴ F (200 µF). The charge is 20 × 10⁻⁶ × 30 = 6.0 × 10⁻⁴ C.