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Formula of Capacitance

Electricity and magnetism formulasScalarClass 12Updated 1 Nov 2026

C=QVC = \frac{Q}{V}
SI unit
F (farad)
Dimensions
[M⁻¹ L⁻² T⁴ I²]
where
  • C capacitance F
  • Q charge on one plate C
  • V potential difference between plates V

The formula of capacitance is C=QVC = \frac{Q}{V}: the charge stored on one plate divided by the potential difference between the plates. It measures how much charge a capacitor holds for every volt, in farads.

For the most common design, two flat plates of area A a distance d apart, capacitance depends only on geometry and the material between the plates: C=Kε0AdC = \frac{K\varepsilon_0A}{d}. Board exams ask for both forms, plus the series and parallel rules and the stored energy 12CV2\frac{1}{2}CV^2.

Quick facts

  • Definition

    C=QVC = \frac{Q}{V}

  • Parallel plate

    C=Kε0AdC = \frac{K\varepsilon_0A}{d}

  • Isolated sphere

    C=4πε0RC = 4\pi\varepsilon_0R

  • SI unit

    farad (F) = C/V

  • Dimensional formula

    [M⁻¹ L⁻² T⁴ I²]

  • Energy stored

    U=12CV2=Q22CU = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

Each symbol in the formula of capacitance

Terms in C = Q / V and C = Kε₀A / d
SymbolMeaningSI unit
CCCapacitanceF
QQCharge on either plate (the plates carry +Q and −Q)C
VVPotential difference between the platesV
KKDielectric constant of the filling, 1 for vacuumno unit
AAArea of overlap of the platesm²
ddGap between the platesm

Capacitance does not depend on Q or V separately. Doubling the voltage doubles the charge, so the ratio stays fixed for a given capacitor. Only the size, shape, spacing and filling can change it.

Where the parallel plate result comes from

Charge ±Q on plates of area A gives a surface density σ=QA\sigma = \frac{Q}{A}. The field between them is E=σε0E = \frac{\sigma}{\varepsilon_0}, and the voltage across the gap is V=EdV = Ed. Then

C=QV=QQε0Ad=ε0AdC = \frac{Q}{V} = \frac{Q}{\frac{Q}{\varepsilon_0A}d} = \frac{\varepsilon_0A}{d}

A dielectric reduces the field by a factor K, so C rises by K. The steps use the surface charge density and the plate field from the electric field page.

Worked examples

Example 1: from charge and voltage

A capacitor stores 60 μC when connected to 12 V. Find its capacitance.

C=60×10−612=5.0×10−6 F=5.0 μFC = \frac{60\times10^{-6}}{12} = 5.0\times10^{-6}\ \text{F} = 5.0\ \mu\text{F}

Example 2: plate size in cm² and gap in mm

Plates of area 200 cm² are 0.50 mm apart in air. Find C, then find it again with a dielectric of K = 4.

A = 200 × 10⁻⁴ = 0.020 m², d = 5.0 × 10⁻⁴ m.

C=(8.854×10−12)(0.020)5.0×10−4≈3.5×10−10 F=354 pFC = \frac{(8.854\times10^{-12})(0.020)}{5.0\times10^{-4}} \approx 3.5\times10^{-10}\ \text{F} = 354\ \text{pF}

With the dielectric: C = 4 × 354 pF ≈ 1.4 nF.

Example 3: combining capacitors

Find the total when 4 μF and 12 μF capacitors are connected in series, and again when they are in parallel.

1Cs=14+112=412⇒Cs=3 μF\frac{1}{C_s} = \frac{1}{4} + \frac{1}{12} = \frac{4}{12}\quad\Rightarrow\quad C_s = 3\ \mu\text{F}

In parallel: C = 4 + 12 = 16 μF.

Series and parallel rules are the reverse of resistors

Combining capacitors
ConnectionRuleWhat is shared
ParallelC=C1+C2+…C = C_1 + C_2 + \dotsSame voltage on each
Series1C=1C1+1C2+…\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \dotsSame charge on each

Resistors add in series and use reciprocals in parallel; capacitors do the opposite. A neat way to remember it: putting capacitors side by side in parallel is like making one capacitor with a bigger plate area, so C adds up. Compare with the resistance formula rules.

Units and dimensions of capacitance

One farad is one coulomb per volt, a very large capacitance. Real values are in μF (10⁻⁶), nF (10⁻⁹) and pF (10⁻¹²). A neat route to the dimensions uses the stored energy U=Q22CU = \frac{Q^2}{2C}, so C is charge squared over energy. Charge squared is I²T². Dividing by energy, M L² T⁻², gives [C] = [M⁻¹ L⁻² T⁴ I²].

Starting from C = Q/V instead leads to the same result, as the dimensional formula of capacitance page shows.

ε₀ has units of F/m as well as C²/(N m²). That follows directly from C = ε₀A/d.

Common slips with capacitors

  • Using the total charge on both plates. Q is the charge on one plate.
  • Forgetting that pF means 10⁻¹² F.
  • Applying the resistor rules to capacitors.
  • Leaving the plate area in cm².

For the energy stored in the field between the plates, see the energy density formula. Every other circuit topic is gathered on the electricity formulas hub for revision.

Frequently asked questions

What is the formula of capacitance?

C = Q / V. For a parallel plate capacitor, C = Kε₀A / d.

Which SI unit is used for capacitance?

The farad (F), equal to one coulomb per volt. Practical capacitors are measured in μF, nF or pF.

Does capacitance depend on voltage?

No. For a given capacitor, charge rises in proportion to voltage, so Q / V stays constant. Only geometry and the dielectric change C.

How does a dielectric change capacitance?

It multiplies the capacitance by its dielectric constant K, because it weakens the field between the plates for the same charge.

What is the formula for energy stored in a capacitor?

U = ½CV², which can also be written as Q² / (2C) or ½QV.