Capacitance and Charge
We saw in the previous tutorials that a Capacitor consists of two parallel
conductive plates (usually a metal) which are prevented from touching each other (separated) by an insulating
material called the "dielectric". We also saw that when a voltage is applied to these plates an electrical
current flows charging up one plate with a positive charge with respect to the supply voltage and the other
plate with an equal and opposite negative charge.
Then, a capacitor has the ability of being able to store an electrical charge Q
(units in Coulombs) of electrons. When a capacitor is fully charged there is a potential difference,
p.d. between its plates, and the larger the area of the plates and/or the smaller the distance between them
(known as separation) the greater will be the charge that the capacitor can hold and the greater will be its
Capacitance.
The Capacitors ability to store this electrical charge ( Q )
between its plates is proportional to the applied voltage, V for a capacitor of known
capacitance in Farads. Capacitance C is always positive and never negative. The greater
the applied voltage the greater will be the charge stored on the plates of the capacitor. Likewise, the smaller the
applied voltage the smaller the charge. Therefore, the actual charge Q on the plates of
the capacitor and can be calculated as:
Charge on a Capacitor
Where: Q (Charge, in Coulombs) = C
(Capacitance, in Farads) x V (Voltage, in Volts)
It is sometimes easier to remember this relationship by using pictures. Here the three quantities
of Q, C and V have been superimposed into
a triangle giving charge at the top with capacitance and voltage at the bottom. This arrangement represents the actual
position of each quantity in the Capacitor Charge formulas.
and transposing the above equation gives us the following combinations of the same equation:
Units of: Q measured in Coulombs, V in volts
and C in Farads.
Then from above we can define the unit of Capacitance as being
a constant of proportionality being equal to the coulomb/volt which is also called a Farad,
unit F. As capacitance represents the capacitors ability (capacity) to store
an electrical charge on its plates we can define one Farad as the "capacitance of a capacitor which
requires a charge of one coulomb to establish a potential difference of one volt between its plates"
as firstly described by Michael Faraday. So the larger the capacitance, the higher is the amount of charge
stored on a capacitor for the same amount of voltage.
The ability of a capacitor to store a charge on its conductive plates gives it its
Capacitance value. Capacitance can also be determined from the dimensions or area,
A of the plates and the properties of the dielectric material between the plates.
A measure of the dielectric material is given by the permittivity, ( ε ),
or the dielectric constant. So another way of expressing the capacitance of a capacitor is;
with Air as its dielectric
with a Solid as its dielectric
where A is the area of the plates in square metres, m2 with the larger the area, the more charge the capacitor can store. d is the distance or separation between the two plates. The smaller is this distance, the higher is the ability of the plates to store charge, since the -ve charge on the -Q charged plate has a greater effect on the +Q charged plate, resulting in more electrons being repelled off of the +Q charged plate, and thus increasing the overall charge. ε0 (epsilon) is the value of the permittivity for air which is 8.84 x 10-12 F/m, and εr is the permittivity of the dielectric medium used between the two plates.
Parallel Plate Capacitor
We have said previously that the capacitance of a parallel plate capacitor is proportional
to the surface area A and inversely proportional to the distance, d
between the two plates and this is true for dielectric medium of air. However, the capacitance value of a capacitor
can be increased by inserting a solid medium in between the conductive plates which has a dielectric constant greater
than that of air.
Typical values of epsilon ε for various commonly used dielectric
materials are: Air = 1.0, Paper = 2.5 - 3.5, Glass = 3 - 10,
Mica = 5 - 7 etc.
The factor by which the dielectric material, or insulator, increases the capacitance of the
capacitor compared to air is known as the Dielectric Constant, (k). "k"
is the ratio of the permittivity of the dielectric medium being used to the permittivity of free space otherwise
known as a vacuum. Therefore, all the capacitance values are related to the permittivity of vacuum. A dielectric
material with a high dielectric constant is a better insulator than a dielectric material with a lower dielectric constant.
Dielectric constant is a dimensionless quantity since it is relative to free space.
Example No1
A parallel plate capacitor consists of two plates with a total surface area of 100 cm2.
What will be the capacitance in pico-Farads, (pF) of the capacitor if the plate separation is 0.2 cm, and the
dielectric medium used is air.
then the value of the capacitor is 44pF.
Charging & Discharging of a Capacitor
Consider the following circuit.
Assume that the capacitor is fully discharged and the switch connected to the capacitor has just been moved to position A. The voltage across the 100uf capacitor is zero at this point and a charging current ( i ) begins to flow charging up the capacitor until the voltage across the plates is equal to the 12v supply voltage. The charging current stops flowing and the capacitor is said to be "fully-charged".
Then, Vc = Vs = 12v.
Once the capacitor is "fully-charged" in theory it will maintain its state of voltage charge
even when the supply voltage has been disconnected as they act as a sort of temporary storage device. However,
while this may be true of an "ideal" capacitor, a real capacitor will slowly discharge itself over a long period
of time due to the internal leakage currents flowing through the dielectric. This is an important point to remember
as large value capacitors connected across high voltage supplies can still maintain a significant amount of charge
even when the supply voltage is switched "OFF".
If the switch was disconnected at this point, the capacitor would maintain its charge indefinitely,
but due to internal leakage currents flowing across its dielectric the capacitor would very slowly begin to discharge
itself as the electrons passed through the dielectric. The time taken for the capacitor to discharge down to 37% of its
supply voltage is known as its Time Constant.
If the switch is now moved from position A to position
B, the fully charged capacitor would start to discharge through the lamp now connected across
it, illuminating the lamp until the capacitor was fully discharged as the element of the lamp has a resistive value. The
brightness of the lamp and the duration of illumination would ultimately depend upon the capacitance value of the capacitor
and the resistance of the lamp (t = C x R). The larger the value of the capacitor the brighter
and longer will be the illumination of the lamp as it could store more charge.
Example No2
Calculate the charge in the above capacitor circuit.
then the charge on the capacitor is 1.2 millicoulombs.
Current through a Capacitor
The current that flows through a capacitor is directly related to the charge on the plates as
current is the rate of flow of charge with respect to time. As the capacitors ability to store charge
(Q) between its plates is proportional to the applied voltage (V),
the relationship between the current and the voltage that is applied to the plates of a capacitor becomes:
Current-Voltage (I-V) Relationship
As the voltage across the plates increases (or decreases) over time, the current flowing through the capacitance deposits (or removes) charge from its plates with the amount of charge being proportional to the applied voltage. Then both the current and voltage applied to a capacitance are functions of time and are denoted by the symbols, i(t) and v(t) However, from the above equation we can also see that if the voltage remains constant, the charge will become constant and therefore the current will be zero!. In other words, no change in voltage, no movement of charge and no flow of current. This is why a capacitor appears to "block" current flow when connected to a steady state DC voltage.
The Farad
We now know that the ability of a capacitor to store a charge gives it its capacitance value
C, which has the unit of the Farad, F. But the farad is an extremely large unit
on its own making it impractical to use, so submultiple's or fractions of the standard Farad unit are used instead.
To get an idea of how big a Farad really is, the surface area of the plates required to produce a capacitor with
a value of one Farad with a reasonable plate separation of just 1mm operating in a vacuum and rearranging the equation
for capacitance above would be:
A = Cd ÷ 8.85pF/m = (1 x 0.001) ÷ 8.85x10-12 = 112,994,350 m2
or 113 million m2 which would be equivalent to a plate
of more than 10 kilometres x 10 kilometres square.
Capacitors which have a value of one Farad or more tend to have a solid dielectric and as "One Farad"
is such a large unit to use, prefixes are used instead in electronic formulas with capacitor values given in micro-Farads
(μF), nano-Farads (nF) and the pico-Farads
(pF). For example:
Sub-units of the Farad
Convert the following capacitance values from a) 22nF to uF, b) 0.2uF to nF, c) 550pF to uF.
a) 22nF = 0.022uF
b) 0.2uF = 200nF
c) 550pF = 0.00055uF
While one Farad is a large value on its own, capacitors are now commonly available with capacitance values of many hundreds of Farads and have names to reflect this of "Supercapacitors" or "Ultracapacitors". These capacitors are electrochemical energy storage devices which utilise a high surface area of their carbon dielectric to deliver much higher energy densities than conventional capacitors and as capacitance is proportional to the surface area of the carbon, the thicker the carbon the more capacitance it has.
Low voltage (from about 3.5V to 5.5V) supercapacitors are capable of storing large amounts of charge due
to their high capacitance values as the energy stored in a capacitor is equal to
1/2(C x V2). Low voltage supercapacitors are commonly used in portable hand
held devices to replace large, expensive and heavy lithium type batteries as they give battery-like storage and discharge
characteristics making them ideal for use as an alternative power source or for memory backup. Supercapacitors used in hand
held devices are usually charged using solar cells fitted to the device.
Ultracapacitor are being developed for use in hybrid electric cars and alternative energy applications to
replace large conventional batteries as well as DC smoothing applications in vehicle audio and video systems. Ultracapacitors
can be recharged quickly and have very high energy storage densities making them ideal for use in electric vehicle
applications.
Energy in a Capacitor
When a capacitor charges up from the power supply connected to it, an electrostatic field is
established which stores energy in the capacitor. The amount of energy in Joules that is stored in this
electrostatic field is equal to the energy the voltage supply exerts to maintain the charge on the plates of
the capacitor and is given by the formula:
so the energy stored in the 100uF capacitor circuit above is calculated as:
The next tutorial in our section about Capacitors, we look at
Capacitor Colour Codes and see the
different ways that the capacitance and voltage values of the capacitor are marked onto its body.
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