Syllabus Sections:-


 2D1   9 Understand the factors influencing the capacitance of a capacitor; area and separation of the plates, permittivity of dielectrics and formula C = K A /d.

Note the wording of the syllabus:- Understand the factors influencing the capacitance...

By saying Understand ......  the formula C = K A /d calculations using the formula can be required.

In the Intermediate Licence course you learned that a capacitor consists of two metal plates separated by an insulating material.

In its simplest form is two metal plates set parallel to each other with a separation gap of insulating material. This material is called a dielectric.

The capacitance is the word used to describe the ability of the two plates to store an electrical charge. The capacitance is proportional to the area of the plates (marked A in the diagram) and inversely proportional to the distance between the plates (marked d in the diagram) and also on the properties of the dielectric between the plates. The dielectric can be as simple as air but other materials used are paper, rubber, polythene, mica and ceramics.

Thus the bigger the area the bigger the charge the capacitor can hold but the bigger the gap the smaller the charge that can be held. So the ideal high value capacitor will have large plates and a very small gap between them.

The Dielectric

The property of the dielectric is called permittivity or the dielectric constant.

It is the dielectric material which determines :-

  • the losses (as heat) which increase as the frequency gets higher (see more below)

  • The variation in the value of the capacitor with a rise in temperature or due to ageing

  • the maximum voltage the capacitor may be used at as known as the working voltage

  • the voltage at which the capacitor would be destroyed also known as the breakdown voltage (see more in the next section).

    The constant K is in fact made up of two parts ... ε0 which is the permittivity of free space (or what you would know as in a vacuum) Using SI units (metric) ε 0 is 8.854 x 10^-12.

    and a second part εr often referred to as k by physicists the dielectric constant of the material called the relative permittivity.

    so K = εo x εr

The formula C = K x A / d

relates the Capacitance (C) to the Dielectric constant (k) to the overlapped area of the plates (A in sq meters)( so if one plate is smaller than the other the size of the area is of the smaller plate) and the distance between the places (d in meters) and gives the answer in farads.

2D3   9  Understand that capacitors have a breakdown voltage and that they need to be used within that voltage.

Capacitors are rated for the maximum continuous voltage that should be applied. Note that the voltage applied will differ as to whether it is DC or AC.

The effect of applying too much PD is to destroy the capacitor and that might occur in spectacular fashion with a small explosion !!!

What actually happens is that the charge on the plates flashes over or makes a hole in the plates. Depending upon the capacity of the capacitor will determine how devastating the destruction will be !!!

2D2   9  Recall that different dielectrics are used for different purposes, e.g. air, ceramic, mica and polyester; and that with some dielectrics, losses increase with increasing frequency.

In addition to ceramic, mica, polyester, and air spaced as the dielectric in capacitors the electrolytic capacitors have a dielectric of a semi liquid conducting compound between their plates which can be of aluminium or tantalum foil. This dielectric is often impregnated paper as a very thin insulating layer.

So what are the losses.

The losses are the loss in the capacitance value of the capacitor which increase with increasing frequency. So what you thought was a capacitor of a certain value when used in a circuit as the frequency of the applied voltage increases the effective value of the capacitance drops.

As has been mentioned before the maximum frequency that a capacitor can operate without loses is determined by the dielectric material.

There are many ways losses can occur in dielectric materials.

A measure of the dielectric efficiency is a ratio of the energy output compared to the energy input of a capacitor.

The most significant loss being dielectric absorption. When the initial inrush of charging current has fallen there follows a small gradual current that appears to soak into the dielectric. Therefore the charge on the capacitor is time dependent. Likewise if the capacitor is now shorted one would think it discharged, but if set aside and shorted a short time later a second discharge can be obtained. Now if the capacitor is charged and discharged at a higher frequency this absorption causes heat to be generated in the dielectric. This heat is a loss of course.

Now it can be shown the higher the frequency of the applied voltage the lower this loss, and it due to the fact that there is less time for the charge absorption into the dielectric. This will now of course mean the usable capacity is now less the higher in frequency we go! So capacitors for HF use need to be selected carefully w.r.t. the dielectric being used.

Dielectric absorption only occurs in solid or liquid dielectrics. EG: ebonite 70%, mica up to 90%, oil nearly 100% if moisture free. The efficiency of air is for most measurements 100% efficient. Air spaced capacitors are very good quality components indeed.

Other losses in dielectrics are conductor loss, chemical action and leakage loss....but none of these need bother you.









Mainly used for variable tuning capacitors capacitance values up to say 1000pf



up to say 300MHz


semi liquid compound

Used for very high value capacitors in Power supplies for smoothing



up to 1MHz


Used in high voltage capacitors



up to 20MHz
Mica This is the best dielectric but is expensive


Very Good

up to and higher than 200MHz


The usual type of capacitor when critical values are not required such as audio coupling and decoupling. Tolerance typically 20%



up to about 100 kHz


High dielectric constant and has good temperature stability



up to 150 - 200MHz

2D7  11  Understand the charging and discharging of a capacitor in a CR circuit and the meaning of the time constant T=CR.

Questions heavy on maths are very likely in this section of the syllabus


Charge and discharge of Capacitors in CR circuits

CR circuits mean a Capacitor and a Resistor linked in series.

In this section we are going to use the maths notation that if two items need to be multiplied together then they are written together without the times (X) symbol. Thus V = I x R would be written as V = IR

Charge Q (measured in Coulombs) on a capacitor

Consider the circuit arrangement shown below

and the following equations:-

Equation A

The charge Q on a capacitor is equal to the Potential Difference  (V) in VOLTS across the capacitor TIMES its capacitance (C) in FARADS

Q = CV

To put some figure on this a capacitor of 1F will store 1C of charge for an applied p.d. of 1V

thus 1C = 1F x 1V

Equation B

The charge in Coulombs, is equal to the current in amps X the time that the current has flowed in SECONDS

Q = Amps x Seconds

Now looking at the diagram above before S1 is closed, VR = zero, VC = zero

Then at the instant S1 is closed, no current has flowed into the capacitor, therefore there is no charge Q and VC = zero.

VR = 12V and by Ohm's law current (I) starts to flow and I = VS / R Amps at that instant.

A short time after S1 is closed a current is flowing, therefore C is charging, VR must be falling because VC + VR = 12V

VR is now VS - VC.

If VR is less then IR is less and the rate of charge of the capacitor falls, until eventually the capacitor is charged to 12V and no current flows then VR = zero.

The graph of VC with respect to time is shown and is known as an EXPONENTIAL CURVE


Time Constant

You will need to have a basic mathematical appreciation of the foregoing paragraphs in order to be able to understand existing circuits involving C and R, and to be able to select values to give correct function in your own circuits.

With Q = Amps x Seconds and Q = CV then

IT = CV thus T = CV/I, from ohm's law R = V/I so T = CR

So the equation is simply :-

T (in seconds) = C (in farads) x R ( in ohms)

where T is the time taken for a capacitor to charge to 66% of the applied voltage VS

T is called the Time Constant, and equally applies to the charge  of a capacitor and the discharge of a charged capacitor via a resistor where similarly discharge by 66% of VS occurs in Time T = CR


How long will it take to charge a capacitor of one microfarad to 66% of the applied voltage, via a resistor of 2 megohms ?

T = CR  

Using the exponential notation

T = 1 x 10-6 x 2 x 106

the 10-6 and 106 cancel each other out and the answer is

1 x 2 = 2 Seconds

T = CR applies to timer circuits like the well known NE555 timer, PSU smoothing circuits, decoupling circuits, delay and de-bounce circuits.

Let's work through another example as it may help you to understand it better.

Whilst under steady state conditions the perfect capacitor presents an infinite resistance to DC voltages however when the capacitor is in a timing circuit the capacitor offers no resistance, it allows current to flow as follows:-

At the instant S1 is closed the current will be 1milliamp (by Ohm's Law I= V/R = 100 /100,000 = 1/1000 - 1 mA

C starts to charge and builds up a voltage.

When the charge on C reaches 50VC, there is 50V left across R and thus the current I is now 50/100,000 = 1/2000 = 0.5mA and thus C is charging slower.

The slow changing voltage curve is called an EXPONENTIAL curve.

At a time T = C (in FARADS ) X R (in OHMS) C =  10 micro F  R = 100k


T = 10 X 10-6 X 100 X 103

T = 101 x 10-6 X 102 X 103

T = 106 X 10-6

T = 1 SECOND, the voltage on C will be ABOUT 2/3 of the initial voltage of 100 V = 66 Volts.

Not in the 2021 Full Syllabus Recall and apply the formulae for calculating the combined values of capacitors in series and in parallel

When capacitors are linked in series the total value can be calculated from the formula:-

The special case is when there are two capacitors linked in series then the simplified formula can be used :-

When capacitors are linked in parallel the total value can be calculated from the formula:-

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