Capacitors

A capacitor or condenser is an electrical or electronic device that can store energy. It stores the energy within the electric field between a pair of conductors (called "plates"). The process of storing energy in the capacitor is known as "charging", and involves electric charges of equal size, but opposite charge, building up on each plate. Capacitors are often used in electric and electronic circuits as energy-storage devices. There are many different types but in physics questions you are usually asked about a simple parallel plate capacitor. In electronics you will learn about the many types and their uses.
	Capacitors are of two types: electrolytic and non-electrolytic. They have two values stamped on them - their capacitance and a working voltage. If the capacitor exceeds this voltage, the insulating layer will break down and the component will short out.
If we connect up circuit A and then close the switch we would observe the bulb lighting up brightly and then getting gradually dimmer until it went out. (This would happen quickly so we would just observe a flash). This is because the brightness of the bulb will depend upon the size of the current flowing. Initially charge would flow quickly onto the plates of the capacitor (brightly lit bulb) then, as the plates began to fill with charge, the rate of charge flow would exponetially decrease (bulb would grow dimmer) until finally the capacitor would be fully charged and no more charge woul flow (bulb would be dark). With circuit B, however we would not even notice that the capacitor was there! the bulb would remain lit all of the time. The capacitor would never be fully charged it would be in the process of charging 50 times a second on opposite plates - therefore there would be a good rate of charge transfer (current) to keep the bulb brightly lit up. The capacitor therefore blocks d.c. current but allows a.c. current through.

Capacitance

The capacitor's capacitance (C) is a measure of the quantity of charge (Q) stored on each plate when a given potential difference or voltage (V) is applied across the plates:

So capacitance is defined as:

The ratio of charge stored on an isolated conductor to the difference in potential.

  Or

The charge required to cause unit potential difference in a conductor

Units

In SI units, a capacitor has a capacitance of one farad (F) when one coulomb (C) of charge is stored when one volt (V) of potential difference is applied across the plates. Since the farad is a very large unit, values of capacitors are usually expressed in microfarads (F), nanofarads (nF), or picofarads (pF).

Parallel Plate Capacitor

For parallel plate capacitors the capacitance (C) is proportional to the area of each of the conducting plates and inversely proportional to the distance between the plates - as long as the area (A)is much, much greater than the distance between the plates (d) squared.

It is also proportional to the permittivity of the dielectric (). The dielectric is the insulator substance that separates the plates - often that is air.

The capacitance of a parallel-plate capacitor is therefore given by:

When there is a difference in electric charge between the plates, an electric field is created in the region between the plates. The electric field that created is proportional to the amount of charge that has been moved from one plate to the other. This electric field creates a potential difference V between the plates of this simple parallel-plate capacitor.

V = Ed

Stored energy

As opposite charges accumulate on the plates of a capacitor due to the separation of charge, a voltage develops across the capacitor due to the electric field of these charges. Ever-increasing efoort must be put in against this ever-increasing electric field as more charge is separated - work has to be done. The energy (measured in joules) stored in a capacitor is equal to the amount of work required to establish the voltage across the capacitor, and therefore the electric field.

We know that W=QV (energy or work done = charge x potenetial difference)

and Q = CV

Let us plot a graph of charge against potential difference

The capacitor is charged with charge Q to a voltage V.

  If we discharged the capacitor by a tiny amount so the potential difference drops, V.  The resulting tiny energy loss (W) can be worked out from the first equation:

W = V × Q      

This is the area of the orange rectangle on the graph.

If we discharge the capacitor completely, we can see that:

Energy loss = area of all the little rectangles

 = area of triangle below the graph

W (for a capacitor charge or dischage)  = ½ QV

OR

Let us plot a graph of potential difference against charge:

The capacitor is charged with charge Q to a voltage V.

  If we discharged the capacitor by a tiny amount of charge, Q.  The resulting tiny energy loss (W) can be worked out from the first equation:

W = V × Q      

This is the area of the orange rectangle on the graph.

If we discharge the capacitor completely, we can see that:

Energy loss = area of all the little rectangles

 = area of triangle below the graph

W (for a capacitor charge or dischage)  = ½ QV

Both methods give us the same outcome!

Now, as Q=CV

W (for a capacitor charge or dischage)  = ½ CV²

or W (for a capacitor charge or dischage)  = ½ Q²/C

On your data sheet you are told that the energy stored is given by:

where V is the voltage across the capacitor - Q is the charge deposited on each plate - you are expected to remember that Q=CV and to work out the other versions of this equation!.

The maximum energy that can be (safely) stored in a capacitor is limited by the maximum electric field that the dielectric can withstand before it breaks down. Therefore, capacitors of the same type have about the same maximum energy density (joules of energy per cubic metre).

Graphical Representation and Quantitative Treatment of Capacitor Discharge

The decay of charge in a capacitor is similar to the decay of a radioactive nuclide. It is exponential decay. If we discharge a capacitor, we find that the charge decreases by half every fixed time interval - just like the radionuclides activity halves every half life.  If it takes time t for the charge to decay to 50 % of its original level, we find that the charge after another t seconds is 25 % of the original (50 % of 50 %).  This time interval is called the half-life of the decay.  The decay curve against time is called an exponential decay. .

The voltage, current, and charge all decay exponentially during the capacitor discharge.

We can perform an experiemnt to obtain the data for us to plot a graph to show this by using the circuit below:

We can charge up the capacitor and then flip the switch and record the voltage and current readings at regular time intervals and plot the data, which gives us the exponential graphs below. 

Let us use a voltage of 12V to charge up a 30F capacitor and have resistance R as 2 M

Note the following about the graphs:

• The half life of the decay is independent of the starting voltage.

• The current  follows exactly the same pattern (as I = V/R)

.

Note that the 'half life' is the same as for voltage

• The charge follows the same pattern, as Q = CV.

The graphs are asymptotic (like the one for radioactive decay) , i.e. in theory the capacitor does not completely discharge but in practice, it does.

The graph is described by the relationship:

Q = Q₀ e ^–t/RC

Compare this to the radioactive decay equation:

the decay constant is equivalent to ¹/_RC

The product RC (capacitance of the capacitor × resistance it is discharging through) in the formula is called the time constant.  The units for the time constant are seconds.

We can show that ohms × farads are seconds.

unit of R = ohms; unit of capacitance = farads

but V=IR so unit of resistance is V/A and C = Q/V so th unit is C/V

unit of product is therefore A/C... but an amp is a coulomb per second so the unit is the second!

  So if we discharge the capacitor for RC seconds, we can easily find out the fraction of charge left:

            V= V₀ e ^–RC/RC = V₀ e ^–1 = 0.37 V₀

So, after RC seconds the voltage is 37 % of the original. 

This fact is used widely by electronic engineers.  To increase the time taken for a discharge we can:

• Increase the resistance.

• Increase the capacitance.

We can link the half-life to the capacitance.  At the half life:

Q = ¹/₂ Q₀

t = t_1/2

Q₀/2 = Q₀ e ^{– t}_1/2^/RC

½ = e ^{– t}_1/2^/RC

2^-1 = e ^{– t}_1/2^/RC

e ^{+ t}_1/2^/RC = 2

ln (2) = t_1/2/RC 

t_1/2  = log_e (2) × RC

t_1/2 = 0.693 × RC

So, the half-life is 69.3 % of the time constant.

Measuring the capacitance 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