# Capacitance: units, formulas, calculation, examples

Capacitance is also defined as the property or capacity of a capacitor or set of electrical capacitors, which is measured by the amount of electrical charge that they can store, separately, per unit of change in electrical potential.

The term capacitance is introduced as a consequence of the creation of an electrical device called a capacitor, invented by the Prussian scientist Ewald Georg von Kleist, in 1745, and independently by the Dutch physicist Pieter van Musschenbroek.

__Capacitor and capacitance__

__Capacitor and capacitance__

A capacitor or capacitor is made up of two conductors that have equal and opposite charges. The conductors are called trusses or plates of the capacitor.

One plate is attached to the positive (+) terminal of a battery, while the other plate is attached to the negative (-). Since the plates have equal and opposite charges, the net charge on a capacitor is zero (0).

Capacitance is the ratio between the charge of a conductor or conductors that form a capacitor and the value of the voltage difference between the plates of the capacitor.

__Units and formulas__

__Units and formulas__

The formula for capacitance is as follows:

C = q / v

Where C is capacitance, *q* the charge (whose unit is the coulomb) and *v* the voltage (volt)

The unit of capacitance is the farad (F), which is equal to coulomb / volt. The farad is a very large unit, so the microfarad (µF) is used, which is equivalent to 10 ^{-6} farad; or the peak farad (pF), which is equivalent to 10 ^{-12} farad.

__How is capacitance calculated?__

__How is capacitance calculated?__

What will be the value of the capacitance of a capacitor whose plates have a charge of 5 · 10 ^{-3} coulombs, and a voltage difference of 6 volts?

C = q / v

= (5 10 ^{-3} coulomb) / (6 volt)

= 8.33 10 ^{-4} farad

__Examples__

__Examples__

The formula for capacitance varies depending on the type of capacitor.

**Parallel Plate Capacitor**

C = kε _{or} A / d

k is the dielectric constant, which has a value of 1 in air and vacuum. For this reason the formula is reduced to:

C = ε _{or} A / d

ε _{o} is the dielectric constant, whose value is close to 8.854 · 10 ^{-12} F · m ^{-1} , A is the area or surface of the parallel plates expressed in m ^{2} , while *d is* the distance that separates the parallel plates.

**Spherical capacitor**

C = 4Πε _{or} R

Where R is the radius of the sphere in meters.

**Concentric sphere capacitor**

C = 4Πε _{or} / (1 / R _{1} – 1 / R _{2} )

**Concentric cylinder capacitor**

C = 2Πε _{or} l / ln (R _{2} / R _{1} )

Where *l* is the length of the concentric cylinders in meters.

__Solved exercises__

__Solved exercises__

**Parallel Flat Plate Capacitor**

What will be the capacity of a capacitor or condenser in air with an area of its plates of 3 cm ^{2} and separated by a distance of 2 mm?

We have the formula:

C = ε _{or} A / d

And the data:

ε _{o} = 8,854 x 10 ^{-12} F · m ^{-1}

A = 3 cm ^{2} (3 · 10 ^{-4} m ^{2} )

d = 2 mm (2 · 10 ^{-3} m)

We simply proceed to replace:

C = (8,854 · 10 ^{-12} F · m ^{-1} ) (3 · 10 ^{-4} m ^{2} ) / (2 · 10 ^{-3} m)

= 1.3281 10 ^{-14} F

**Capacitor or sphere-shaped capacitor**

If we consider the Earth as a spherical capacitor with a radius (R) of 6,370 km: What will be the value of its capacitance?

Data:

C = 4Πε _{or} R

Π = 3.1416

ε _{o} = 8,854 10 ^{-12} Fm ^{-1}

R = 6,370 Km (6.37 · 10 ^{6} m)

We proceed again to substitute the values in the capacitance formula:

C = (4 3.1416) (8.854 10 ^{-12} F m ^{-1} ) (6.37 10 ^{6} m)

= 7.09 10 ^{-8} F

= 709 µF

__Capacitor combination__

__Capacitor combination__

Capacitors or capacitors can be combined in series or in parallel.

**Capacitors in series**

The image above shows three capacitors in series (C _{1} , C _{2} and C _{3} ), as well as a battery with its positive (+) and negative (-) terminals. These capacitors present a series of characteristics in relation to their voltage, charge and capacitance.

**Voltage drop (ΔV) across capacitors**

ΔV _{t} = ΔV _{1} + ΔV _{2} + ΔV _{3}

The total voltage drop across a set of series capacitors is equal to the sum of the voltage drops across the capacitors.

#### **Capacitor ****charging**

Q _{t} = Q _{1} = Q _{2} = Q _{3}

The same amount of charge circulates through the capacitors arranged in series.

**Capacitance of capacitors**

The equivalent capacitance of series capacitors has the following relationship:

1 / C _{eq} = 1 / C _{1} + 1 / C _{2} + 1 / C _{3}

**Capacitors in parallel**

Above we have three capacitors arranged in parallel (C _{1} , C _{2} and C _{3} ), which have the following behavior in relation to the voltage drop, the charge and the capacitance:

**Voltage drop across capacitors**

ΔV _{t} = ΔV _{1} = ΔV _{2} = ΔV _{3}

In parallel capacitors, the total voltage drop across the capacitors is the same as that for each of the capacitors.

**Capacitor charging**

Q _{t} = Q _{1} + Q _{2} + Q _{3}

In a parallel system the total charge on the capacitors is equal to the sum of the charge on all the capacitors.

**Capacitance of capacitors**

C _{eq} = C _{1} + C _{2 }+ C _{3}

In a parallel system the equivalent capacitance of them is equal to the sum of the capacitances of all the capacitors.

**Example of an exercise**

A schematic of three capacitors is shown above: C _{1} and C _{2} are arranged in series and they are in parallel with C _{3} . The capacitance of the capacitors are as follows: C _{1} = 5 µF, C _{2} = 6 µF and C _{3} = 3 µF. Find the equivalent capacitance of the circuit.

First find the equivalent capacitance of C _{1} and C _{2} that are in series.

1 / C _{eq1,2} = 1 / C _{1} + 1 / C _{2}

1 / C _{eq1,2} = 1/5 µF + 1/6 µF

1 / C _{eq1.2} = ( _{11/30} ) µF

C _{eq1.2} = 30 µF / 11

= 2.72 µF

Capacitors 1 and 2 are in parallel with C _{3} . So the equivalent capacitance of C _{1} , C _{2,} and C _{3} is equal to C _{eq1,2} + C _{3} .

C _{eq1,2,3} = 2.72 µF + 3 µF

= 5.72 µF