Ch 2 · 19 Capacitance · A Level Physics 9702
Topic 2 of 2 in Electric Fields and Capacitance — A Level Physics 9702 — 42 lessons.
Introduction
A camera flash charges up for a few seconds, then dumps its energy in a thousandth of a second. The device that stores that charge is a capacitor. This topic shows how much charge and energy capacitors store and how they discharge. (Syllabus topic 19: sections 19.1–19.3.)
Key ideas — 19.1 (a) Capacitance
Capacitance is the charge stored per unit potential difference:
🧑🏫 Teacher's explanation
Think of a capacitor like a water tank. The capacitance C is essentially the 'width' or 'capacity' of the tank. If you want to raise the water level (the potential difference V) by a certain amount, a wider tank (higher C) will hold much more water (charge Q) than a narrow one.
📖 Definitions
🔍 Worked example — Standard
Two capacitors, C_1 = 10muF and C_2 = 20muF, are connected in parallel. Calculate the total capacitance C_total.
🔍 Worked example — Exam-style
An isolated spherical conductor has a radius of 5.0 cm. Calculate its capacitance. (Use varepsilon_0 = 8.85 × 10^-12 F m^-1)
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 19.1 (b) Capacitors in parallel and in series
Parallel: each capacitor has the same p.d. V; the total charge is the sum of the charges: Q = Q_1 + Q_2 + … = C_1V + C_2V + …. Dividing by V:
🧑🏫 Teacher's explanation
Think of a capacitor like a water tank. Capacitance is essentially the 'capacity' of the tank to hold charge. When you connect capacitors in parallel, it is like adding more tanks side-by-side; you have more total space to store charge for the same pressure (voltage), so the total capacitance increases. It's just like adding more pipes in parallel to increase water flow.
📖 Definitions
🔍 Worked example — Easy
Two capacitors, $C_1 = 2.0\text{ $\muF and $C_2 = 3.0\text{ $\muF, are connected in parallel. Calculate the total capacitance C_total.
🔍 Worked example — Standard
Two capacitors, $C_1 = 4.0\text{ $\muF and $C_2 = 12.0\text{ $\muF, are connected in series. Calculate the total capacitance C_total.
🔍 Worked example — Exam-style
A $10\text{ $\muF capacitor and a $20\text{ $\muF capacitor are connected in series across a 60 V supply. Calculate the charge Q stored on the $10\text{ $\muF capacitor.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 19.2 Energy stored
As a capacitor charges, each extra bit of charge must be pushed onto the plates against the increasing p.d. On a graph of p.d. against charge (a straight line through the origin, gradient 1/C), the area under the graph is the energy stored:
🧑🏫 Teacher's explanation
Think of a capacitor like a compressed spring. When you push the plates together (or push charge onto them), you are doing work against an electric force. The more charge you add, the harder it becomes to add even more because the existing charge is repelling the incoming charge. This 'effort' you exert is stored as potential energy.
📖 Definitions
🔍 Worked example — Easy
A capacitor has a capacitance of 470 pF. Calculate the energy stored when the potential difference across it is 12.0 V.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 19.3 Discharging through a resistor
When a charged capacitor discharges through a resistor R, the current is large at first (the p.d. is large) and decreases as the charge, and so the p.d., falls. The decay is exponential:
🧑🏫 Teacher's explanation
Imagine a large water tank connected to a narrow pipe. When the valve is opened, the water flows out very quickly because the pressure (potential difference) is high. As the water level drops, the pressure decreases, and the flow rate (current) slows down. The discharge of a capacitor is exactly like this, but instead of water, we are moving electric charge.
📖 Definitions
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Example 1. A 470 μF capacitor is charged to 12 V. Find the charge and energy stored.
🔍 Worked example — Standard
Two capacitors, 2.0 μF and 3.0 μF, are connected (a) in parallel and (b) in series. Calculate the total capacitance for each case.
🔍 Worked example — Exam-style
A 470 μF capacitor is charged to 12 V and then discharged through a 10 kOmega resistor. Calculate the potential difference across the capacitor after 3.0 s.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) Charge stored per unit potential difference: C = Q/V. (2) Same V across each; Q = Q_1 + Q_2 = C_1V + C_2V; so C = C_1 + C_2. (3) C = 2W/V^2 = 1.0/10^4 = 1.0 × 10^-4 F = 100 μF. (4) tau = 47 × 10^3 × 220 × 10^-6 = 10.3 s; e^-2 = 0.14 (14%). (5) As charge leaves the plates, the p.d. across the capacitor falls, so the p.d. across the resistor and hence the current (I = V/R) fall.