Electricity and Magnetism



In the following set-up, two strips of polythene have been rubbed with wool, and when bought close together, they repel each other:

The polythene has become 'electrostatically charged' by friction with the wool.

If a strip of Perspex  is rubbed with wool, and then bought close to the suspended polythene strip, they attract each other.

This indicates that there are two types of charge, and we call these positive and negative. The polythene strips are taken as having a negative charge, and the Perspex a positive charge.

Observations like those above indicate that:

So, in brief: We can interpret the above observations in terms of atoms: The SI unit of charge is the coulomb (C)


The above represents the situation when the forces are attractive. No subscripts are needed on the Fs, since the force on each charge is the same in magnitude, but they are opposite in direction (recall Newton’s third law). According to Coulomb's law:

Note: For two charged spheres, r is the distance between their centres.


Notice how similar the above is to the gravitational case. However, one difference is that in the gravitational case, the force between two masses is always attractive, whereas in the electric case the force can be attractive or repulsive, depending on the signs of the charges.

Also, in the gravitational case, the constant of proportionality is the universal gravitational constant G, and its value does not depend upon the nature of the medium between two masses. However, in the electric case the force between two charges does depends upon what is between them.

To take the medium into account we write Coulomb’s law as:

A more commonly used, and equivalent unit, is ‘farads per metre’ (F m-1)


There is no doubt that the forces between charges are real, since we can observe the effect of such forces, as in the simple experiment with charged strips above. However, there is no apparent physical contact between the strips, and we cannot see what produces the forces. To ‘explain’ such ‘action at a distance’, we assume that an electric charge is surrounded by an invisible 'electric field', and that it is the interaction between fields that produces the observed effects.

An electric field is any region where an electric force may be experienced. We represent such fields by lines with arrows on them.

The direction of the field at a point, represented by an arrow, is defined as the direction of the force on a positive charge at that point. Thus, arrows point away from a positive charge and towards a negative charge.



Thus, if a small positive point charge q is placed at a point in an electric field, and it experiences a force F, then the electric field strength E at that point is defined by:
E is a vector with the same direction as F (the force on the positive charge, q).
From the definition, the unit of E = unit of (F/q) = N C-1.
A more commonly used, but equivalent, unit of E is volts per metre (V m-1).

E due to a point charge

Consider a small positive test charge q placed at P, a distance r from a point charge +Q:

By Coulomb’s law, the force F on q is:

In the diagram, the point charge is represented as being positive, but the above equation gives E for both a positive or negative point charge Q.

E due to a charged spherical conductor

So far as external effects are concerned, we can treat a spherical conductor, having charge Q distributed uniformly over its surface, as if all the charge were at its centre.

Thus, if its radius is a, the field strength E at a distance r from its centre (r ³ a):




Potential is a scalar quantity, and the potential at a point due to a number of charges is the algebraic sum of the potentials at that point due to each charge.

Potential due to an isolated point charge

We can show that the electric potential VE, a distance r from charge Q is given by:
And also that:
(Hence, the unit V/m for electric field strength, E.)
Potential due to a charged conducting sphere
It can be shown that:

i) The potential V at a distance r from the centre is given by:

Where, r ³ a, i.e. on the surface and outside the sphere.

ii) Inside the sphere the potential is the same at all points and the same as on the surface. Thus,

Potential difference (‘p.d.’) So, p.d. equals 'work done per unit charge' = work/charge.

Thus, the work W done in moving a charge Q through a p.d. V is given by:

Note: The work done in taking a charge around a closed loop is zero - since it arrives back at its original point, and so the p.d. between its start and end points is zero.

Parallel plates

A battery is applied to two flat, parallel metal plates and produces an electric field as represented above.

Near the edges the field in not uniform, and so is not given accurately by this equation.


In an isolated conductor there can be no differences in potential, since these would set up potential gradients (º electric fields), and charges would redistribute themselves until they had destroyed the field.

Any surface or volume over which the potential is constant is called an equipotential. The surface or volume may be that of a material body or a surface or a volume in space.

Since the change in potential between any two points in an equipotential surface is zero, there is no potential gradient, so there is no electric field. Electric field lines are therefore at right angles to equipotential surfaces, since the field lines then do not have components in the surface.

a) All spheres centred on a point charge are equipotential surfaces:

b) The potential gradient between two parallel metal plates is constant (apart from at the edges) and all planes parallel to and between the plates are equipotential surfaces:
Electric field strength, E = 3V/3cm = 1 V/cm.

The above represents the basic structure of a capacitor.



A capacitor is a device that can store electric charge. It is basically a very simple device consisting of two metal sheets, separated by an insulating material. Often, in practical capacitors, the sheets are rolled up, so the capacitor becomes cylindrical, and is similar to a roly-poly pudding in cross-section.

The above represents a capacitor made of two metal plates, with a battery applied across them. There is a current while the capacitor is ‘charging up’ - electrons flow from one plate to the other. When charging is complete, the p.d. across the capacitor equals that of the battery.

When charged up, an electric field exists between the plates. The direction of the field is defined as that of the force on a positive charge placed between the plates. If charge q were between the plates and experienced a force F, then the magnitude of the electric field, denoted by E.

From this we get the more commonly used, but equivalent, unit for E of volts per metre (Vm-1).

If the voltage across a capacitor is too great, the insulator breaks down, and becomes a conductor. This can make the capacitor get hot or even explode. The working voltage of a capacitor is normally written on it, indicating the maximum voltage that can be safely applied to it.


The capacitance C of a capacitor is a measure of its ability to store charge. By definition:

From the definition, the unit of capacitance is C/V or CV-1 ('coulombs per volt'), which we call the farad, F:   i.e. 1F = 1CV-1.

Energy stored in a capacitor

A graph of p.d. versus charge is a straight line through the origin:


  1. find the charge on the capacitor
  2. find the energy stored using each of the above equations


Capacitors in parallel

Key facts:
  1. There is the same p.d. V across each capacitor
  2. The total charge stored, Q = Q1 + Q2 + Q3
We want to find the single capacitor equivalent to all three, i.e. what capacitor C would store the same charge as the above three if the same p.d. were applied across it?

C may be referred to as the equivalent or effective or combined capacitance.

By fact (1) we do not need a subscript on V since the p.d. is the same for each.

And, by fact (2):

Notice that this expression is similar to that for resistors in series.

Capacitors in series

Key facts:
  1. There is the same charge on each capacitor
  2. V = V1 + V2 + V3
What single capacitor C would carry the charge Q if the p.d. V were applied across it?

By fact (1) we do not need a subscript on Q since the charge is the same for each.

And, by fact (2):

Notice that this expression is similar to that for resistors in parallel.


Find the single capacitor equivalent to:


Find the single capacitor equivalent to:


Find the single capacitor equivalent to:

For the capacitors in series,

Notice that V1+V2= 100V, the applied voltage.

Joining two capacitors

The two capacitors are initially separate. When joined (the dotted lines):
  1. The total capacitance, C = C1 + C2
  2. The capacitors acquire the same p.d.
  3. The total charge remains constant

In (a), capacitor C1 has been charged by a 60V supply.
In (b), C1 has been joined across an uncharged capacitor C2.

We see that the final amount of stored energy is less than the initial amount. When the capacitors are joined, current has to flow to achieve the redistribution of charge, and this generates some heat loss.


With S in position 1, the capacitor will charge up to its maximum value of Q0 = CV0.
When S is moved to position 2, the capacitor will start to discharge through R:

a) Charge

The charge starts at its maximum value and then falls to zero:

   The 'decay' is referred to as being 'exponential', the variation of Q with time obeying the equation:

Where 'e' is a number with the value 2.718 ........... It is the base of 'natural logarithms'.

The magnitude of the power that e is raised to, (t/CR), has no unit. So, since time t is in seconds, then CR must also be in seconds, in order that the units on top and bottom cancel. Thus, CR has the unit of time, and is called the ‘time constant’ of the circuit - it determines the rate at which Q approaches zero:

Half-life, T1/2

Theoretically, the above curves never reach the time axis, since time t has to equal ‘infinity’ for Q to equal zero. Hence, to compare different CR circuits, we introduce the idea of  'half life' of a CR circuit. The term is analogous to the same term used in radioactive decay.

Relation between time constant and half-life

To follow the derivation below you may have to read up about 'logarithms' - your syllabus may not require you to know the derivation - but the final relationship, between CR and T1/2, is important:

Thus, the capacitor is half discharged when the time reaches 69% of the time constant.

While the capacitor is discharging:

The half-life is the same for Q, I and V - i.e. the time T1/2, given by the last equation, is the time for: The value of T1/2 can be read directly from a Q-t graph (as above), or a I-t or V-t graphs (as below), and hence the value of the time constant, CR, can then be found using the above equation.

b) Current

Current equals rate of flow of charge. Thus, the slope (or gradient) of the Q-t graph at any time equals the current at that time:

c) Potential difference

The equation for this comes directly from that for Q, since V = Q/C


a) Charge

When the switch S is closed, the charge on the capacitor rises from zero to its maximum value of Q0 = CV0. The variation of charge Q with time t has the form:

The charge Q on the capacitor at time t seconds after the switch is closed is given by the equation:
We again use the idea of half-life. Its meaning in this case is indicated by the above graph for charge and the graphs below for current and voltage. The value of half-life is the same in each case, and found from the same equation as before:
b) Current

We can show that:

c) Potential difference

The equation for this comes directly from that for Q, since V = Q/C


A Level Physics - Copyright © A C Haynes 1999 & 2004