Capacitor Tutorial 3 - Physics of Capacitors

If we pump electrons onto the negative plate, electrons are repelled from the negative plate.  Since positives do not move, a positive charge is induced.  The higher the potential difference, the more charge is crowded onto the negative plate and the more electrons repelled from the positive plate.  Therefore charge is stored.  The plates have a certain capacitance.


Capacitance is defined as:


Since we have a positive and negative plate, we have an electric field.  Capacitance is measured in units called farads (F) of which the definition is:

1 Farad is the capacitance of a conductor, which has potential difference of 1 volt when it carries a charge of 1 coulomb.

  So we can write from this definition:

            Capacitance (F) = Charge (C)

                                     Voltage (V)

  In code, this is written:

 C = Q


  [Q - charge in coulombs (C); C – capacitance in farads (F); V - potential difference in volts (V)]


A 1 farad capacitor is actually a very big capacitor indeed so instead we use microfarads (mF) where 1 mF = 1 × 10-6 F.  Smaller capacitors are measured in nanofarads (nF), 10-9 F, or picofarads (pF), 1 × 10-12 F.  A working voltage is also given.  If the capacitor exceeds this voltage, the insulating layer will break down and the component shorts out.  The working voltage can be as low as 16 volts, or as high as 1000 V.

The voltage rises as we charge up a capacitor, and falls as the capacitor discharges.  The current falls from a high value as the capacitor charges up, and falls as it discharges.  We can see this in the graph below.

Why does a capacitor allow AC to flow?

Connect a capacitor as shown:

  If we connect a capacitor in series with a bulb:

We can say that a capacitor blocks d.c., but allows a.c. to flow.  On the forward half-cycle the capacitor is charging up.  As the current passes through the bulb, the filament lights.  On the reverse half cycle, the capacitor discharges, and the bulb lights up.  In electronic circuits the capacitor acts as a filter to block DC.


Measuring Capacitance

This circuit can be used to measure the value of a capacitor:

We can use I = Q/t to work out the charge going onto the plates.  We also know that f = 1/t, so we can combine the two relationships to give I = Qf, therefore Q = I/f

Since C = Q/V, we can now write C = I/fV



A capacitor is connected to a 12-volt power supply by a reed switch operating at 400 Hz.  The ammeter reads 45 mA.  What is the capacitance of the capacitor?

            C = 0.045 A × (400 Hz × 12.0 V) = 9.38 × 10-6 F = 9.38 mF

  You can give your answers in farads or microfarads, as long as you are consistent.


Capacitance of a Capacitor

  We can show that the capacitance of a capacitor depends:

  So we can write:


µ A/d


C = kA/d

This means that we can increase the capacitance by increasing the area of the plates or by reducing the gap between the plates.  However this will limit the voltage we can apply across the plates, as the dielectric may break down and sparks jump across the gap.

The proportionality constant is called the permittivity of free space.  It is given the physics code e0 (“epsilon nought”; epsilon is a Greek letter ‘e’). 

It has the numerical value 8.85 × 10-12 C2 N-1 m-2. 

We can therefore rewrite the formula as:

 C = e0 A/d.


A 1 farad capacitor has its plates separated by 1 mm of air.  What is the area of its plates?

            Formula first: C = e0 A/d therefore A = Cd/e0

            Put the numbers in: A = (1 F × 1 × 10-3 m) × 8.85 × 10-12 C2 N-1 m-2

                                                = 1.12 × 108 m2

  This would give us a capacitor with plates 10 km × 10 km, which is rather impractical.  A farad is a very big unit, and we are much more likely to use microfarads (mF) or nanofarads (nF).


The insulating gap between the plates of a capacitor is called the dielectric.  The reference dielectric is a vacuum, but air gives a value that is very similar.  We can use a dielectric other than air.  Some insulating materials do not affect the capacitance of the capacitor at all, but there are others, for example polythene or waxed paper that make the capacitance rise quite a lot.  This happens because the molecules become polarised, which means that the electrons move slightly towards the positive plate, leaving a deficiency of electrons, hence a positive charge, at the other end.  We see this:


 The presence of the polarised molecules alters the electric field between the plates.  Electric field goes from positive to negative. 


The relative permittivity or the dielectric constant tells us how much the capacitance of a capacitor is increased relative a vacuum or air.  It is given the physics code er ("epsilon-r"), and has no units.  Our relationship gets modified to:

C = e0er A/d

The table shows some typical values of dielectric constant.  Water is not a good practical dielectric as the impurities dissolved in it make it conduct.



Dielectric constant









Waxed paper




Water (pure)


Barium titanate



Practical Capacitors

Very few capacitors consist of flat plates that we have looked at so far.  Instead, they consist of two layers of aluminium foil alternating between two layers of dielectric.  The whole lot is rolled up like a Swiss roll to make a compact shape.

Non electrolytic capacitors have a mica or polyester dielectric.  The value of the capacitors made in this way is quite low, up to about 10 mF.

  Electrolytic capacitors are capable of holding a much bigger charge.  The aluminium metal plates are either side of a sheet of paper soaked in aluminium borate.  When the capacitor is charged up, there is a chemical reaction that deposits an aluminium oxide layer on the positive plate.  This acts as the dielectric.  The electrolyte soaked paper acts as the negative plate.


Why does a capacitor discharge exponentially?

Think about a capacitor that is fully charged.  Let's suppose we remove 1 % of its charge.  Its charge goes down to 99 %.  The new charge is 99 % of the original

Now let's remove another 1 % of the charge that is left, i.e. the 99 % of the original.  1 % of 99 % = 0.99 %.  New charge is now 98.01 %.  And so on...

Now suppose we lose that fraction in each unit time.  We can say that the capacitor discharges a constant fraction of what is left in each unit time.  The charge lost in each unit of time can be worked out as:

Charge loss per unit time = change in charge ÷ time interval

This is the rate of change of charge which is also the current.  So we can write this as an equation:

From Ohm's Law (V = IR) we can also write:

The minus sign tells us that there is a discharge going on.

From basic capacitance we know that Q = CV.  Therefore we can write:

The term RC is the time constant, which we have seen in the previous tutorial.  Remember that ohms × farads = seconds.  We can rewrite the expression as:


In mathematics one kind of differential equation is one in which a quantity, x, decays with a constant fraction for each unit time.  An alternative is to say that the rate of decrease in x is proportional to the amount of x that remains.  The differential equation is in the form:

The constant, k, is the fraction of the the quantity x that decays every second.

This is a differential equation in which 1/RC is the constant.

The solution to this can be derived using the calculus process of integration.  This gives the equation:

We can take natural logarithms to rewrite this as:

If we plot ln V against time, we get a straight line graph of the form y = mx + c.

Notice the following: