Topic 6 of 7
Capacitance and plate charge
A capacitor stores charge separation between conductors. Its capacitance uses the charge magnitude on either plate, even when the two-plate system has zero net charge.
Define charge and potential difference locally
A capacitor has separated conductors that acquire opposite charges. For the usual initially neutral pair, one plate has +Q and the other -Q. Let V be the magnitude of their potential difference.
Capacitance C is charge stored divided by potential difference. Q is the magnitude on either plate, not the pair's algebraic net charge and not the sum of both magnitudes. The SI unit is the farad, F:
= kg-1 m-2 s4 A2
The C in the unit ratio is the coulomb; the quantity C is capacitance. Since a coulomb is A s and a volt is kg m2 s-3 A-1, their ratio gives these base units. F as a unit means farad, while F as a force symbol has unit newton.
Useful conversions are 1 microfarad = 10-6 F, 1 nanofarad = 10-9 F and 1 microcoulomb = 10-6 C. A rough 100 microcoulomb at 10 V suggests C of order 10 microfarads.
A linear capacitor has constant capacitance
For a given linear capacitor within its operating range, Q is proportional to V. C is then constant. This is a device-model property, not a guarantee that arbitrary Q/V pairs from every device agree.
Worked plate charge
10.0 microfarads at 12.0 V
= 120 × 10-6 C = 120 microcoulombs
The plates carry +120 and -120 microcoulombs. Their net charge is zero, but separated charge and a potential difference remain.
Capacitance uses either plate's charge magnitude
For the supplied linear 10.0 µF capacitor, C = Q/V. The plates carry opposite signs, while Q in this charging account is the magnitude on one plate.
A charged pair can have zero net charge
Q = 120 µC for C = Q/V. Adding +120 and -120 µC gives the pair's zero net charge; adding their magnitudes does not give the Q used in this formula.
Q against V has gradient C
The gradient is 80 µC / 8.0 V = 10.0 µF. Both Q/V and this constant gradient describe the stated linear capacitor. Do not infer a constant capacitance for an arbitrary device without that condition.
= 10.0 × 10-6 C/V = 10.0 microfarads
The model points at V = 0, 4, 8 and 12 V have Q = 0, 40, 80 and 120 microcoulombs. A steeper Q-against-V line corresponds to a larger capacitance.
Optional check An ideal 10.0 microfarad capacitor has a potential difference of 12.0 V. What charge description belongs with C = Q/V?
For a measured record, identify how charge and voltage were determined, retain their units and resolutions, and examine whether the relation is linear over the tested range. The plotted values here are supplied model values, not experimental measurements. Reversing the graph axes changes the gradient to 1/C, as used in the energy calculation.