9478 / 2027
Electric Fields overview

Lesson 6 of 7

Capacitance and plate charge

  • I can define capacitance as charge stored divided by potential difference and apply C = Q/V.Syllabus

    Syllabus 9478, 14(j). Define capacitance as charge stored divided by potential difference and apply C = Q/V.

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.

C = Q/V

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:

1 F = 1 C/V
= 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

Q = CV = (10.0 × 10-6)(12.0)
= 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

Opposite 120-microcoulomb plate charges correspond to twelve volts across ten microfaradsTwo separated conducting plates form a capacitor with supplied capacitance ten microfarads. The left plate has charge positive 120 microcoulombs and the right plate negative 120 microcoulombs. A dimension-like voltage bracket labels twelve volts between the plates; it is a potential difference, not a distance measurement or wire. Four plus and four minus symbols are qualitative charge-sign markers, not literal particle counts. The pair's total charge is zero while each plate has a nonzero charge magnitude. Plate dimensions and spacing are schematic; no geometrical capacitance formula is implied. No conducting current is drawn across the gap.C = 10.0 µF+120 µC-120 µC+-+-+-+-12.0 V between plates

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 charge-voltage gradient is ten microcoulombs per voltVoltage is horizontal in volts, charge magnitude vertical in microcoulombs. The supplied linear-capacitor model passes through zero, then four volts and forty microcoulombs, eight volts and eighty microcoulombs, and twelve volts and 120 microcoulombs. Filled circles identify the four model pairs. A slope triangle from the four-volt point to the twelve-volt point has horizontal increase eight volts and vertical increase eighty microcoulombs. Its gradient is ten microcoulombs per volt, equivalent to ten microfarads. It is this stated device model that makes the gradient constant; no measurement or universal linearity of every component is asserted.04812040801208.0 V80µCQ / µCV / V

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.

The plate labels give opposite charges of equal magnitude. The separate Q-against-V graph is a supplied linear model with Q on the vertical axis; its gradient is capacitance.
Q-V gradient = (120 × 10-6)/12.0
= 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.

Try it step by step

Work it out in steps: Charge Q = CV in microcoulombs, then Energy stored 1/2 C V2 in millijoules.

  1. Problem 1. A 10 microfarad capacitor is charged to 12 V. Find the charge stored and the energy stored.
  2. Problem 2. A 47 microfarad capacitor is charged to 9.0 V. Find the charge stored and the energy stored.
  3. Problem 3. A 220 microfarad capacitor is charged to 5.0 V. Find the charge stored and the energy stored.
Show the answer

Problem 1.

Charge Q = CV in microcoulombs: Q = 10 x 10-6 x 12 = 1.2 x 10-4 C = 120 microcoulombs

Energy stored 1/2 C V2 in millijoules: E = 1/2 x 10 x 10-6 x 122 = 7.2 x 10-4 J = 0.72 mJ

Problem 2.

Charge Q = CV in microcoulombs: Q = 47 x 10-6 x 9.0 = 4.23 x 10-4 C = 423 microcoulombs

Energy stored 1/2 C V2 in millijoules: E = 1/2 x 47 x 10-6 x 9.02 = 1.90 x 10-3 J = 1.90 mJ (3 sf)

Problem 3.

Charge Q = CV in microcoulombs: Q = 220 x 10-6 x 5.0 = 1.1 x 10-3 C = 1100 microcoulombs

Energy stored 1/2 C V2 in millijoules: E = 1/2 x 220 x 10-6 x 5.02 = 2.75 x 10-3 J = 2.75 mJ (3 sf)

Check your understanding

An ideal 10.0 microfarad capacitor has a potential difference of 12.0 V. What charge description belongs with C = Q/V?

  1. Q is zero because the two plate charges cancel algebraically.
  2. Q is 120 microcoulomb; the plates hold +120 and -120 microcoulomb.
  3. Q is 240 microcoulomb, adding both plate-charge magnitudes.
Show the answer

Q is 120 microcoulomb; the plates hold +120 and -120 microcoulomb.

Q = CV = 120 microcoulomb is the magnitude on one plate. Opposite plate charges give a neutral pair while preserving charge separation and stored energy.

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.