In capacitor problems first find what is known and what must be found, and make the units consistent (µF, nF, pF → F; µC → C). The basic relation is C = q/U, from which q = C·U and U = q/C follow. In parallel the voltage is the same and the total capacitance is C = C₁ + C₂ + …; in series each capacitor carries the same charge, the voltages add, and 1/C = 1/C₁ + 1/C₂ + …. In a series chain the voltage is shared inversely to capacitance: the capacitor with the smaller capacitance has the larger voltage. For a flat capacitor use C = εε₀S/d and remember to convert millimetres to metres. At the end, judge whether the size of your answer is reasonable.
Worked examples
2 µF and 3 µF in parallel, U = 10 V. C = 2 + 3 = 5 µF; q = C·U = 5 · 10 = 50 µC.
6 µF and 3 µF in series, 90 V in total. C = 6 · 3 : 9 = 2 µF; q = 2 · 90 = 180 µC (on each). U₁ = 180 : 6 = 30 V, U₂ = 180 : 3 = 60 V; 30 + 60 = 90 V.
Class activity
Make your own problem: choose two capacitances and a source voltage and find the voltage on each in a series connection. Check in your notebook: the voltages must add up to the source voltage.
Practice
1
What is the same in capacitors connected in series?
The charge on each capacitor.
2
A capacitor C = 8 µF is charged to 25 V. The charge in µC?
200
3
3 µF and 9 µF in parallel are connected to 20 V. The total charge in µC?
240
4
Why does a smaller-capacitance capacitor get the larger voltage in a series chain?
The charge is the same and U = q/C, so smaller C gives larger U.