This episode introduces fields, field lines and equipotentials in the context of electric fields. Much of this is revision from g-fields, but with the slight added twist of needing to take account of the sign of charge when examining electric fields.

Lesson Summary

• Discussion and demonstration: Field lines (20 minutes)
• Discussion and student experiment: Equipotentials (35 minutes)
• Student questions: Field lines and equipotentials (10 minutes)

Discussion and demonstrations: Field lines

Electrical charges exert forces upon one another. Just as with gravity, these forces can be understood in terms of fields that exist between charged particles.

What is the basic law of force between charges? (Like charges repel, unlike charges attract.)

How do these forces occur if, as usual, the charges are not in contact? (The presence of electric fields between the charges. A field is set up by a charge, and any other charge in that field will experience a force due to the field.)

How do we usually represent these fields? (With field lines or lines of force.)

Now we have to be careful with our use of field lines to represent electrical fields. In the gravitational field all forces are attractive and so putting a direction on the field line is unambiguous – it gives the direction a mass will feel a force at a point in a field. Because both attraction AND repulsion can occur in an electric field, we introduce the following convention that is consistent with the fact that only one type of mass exists, i.e. positive mass:

The direction of a field line in the electric field is the direction of the force on a small positive charge.

Thus if a positive charge is placed at a point in a field, it will feel a force in the direction of the field line at that point, but if a negative charge is placed there, it will feel a force in the opposite direction to the direction of the field line at that point.

We can see the field lines for certain geometries of charge using a simple demonstration with semolina powder in oil, between two electrodes connected to a high voltage supply.

Episode 406-1: Demonstration – electric field lines (Word, 74 KB)

This allows us to visualise the following fields:

Again, there are some basic rules and observations about field lines:

• They never start or stop in empty space – they stop or start either on a charge or at infinity.
• They never cross – if they did, a small positive charge placed there would feel forces in different directions, which could be resolved into the one true direction of the field line there.
• The density of field lines on a diagram is indicative of the strength of the field.
• The second diagram above also shows a point exactly between two like charges where no field exists (since the forces on a charge placed there would be exactly equal and opposite in direction). Such a point is called a neutral point.

Discussion and student experiment: Equipotentials

Exactly as with the gravitational field, we define an equipotential surface as one that joins points of equal potential in the field; in other words, no work is done in moving a charge on an equipotential surface. Although our discussion and definition of potential will be almost exactly the same as in the gravitational case, we will leave that to a later episode. For now, we just need to know that:

• Equipotential surfaces are perpendicular to field lines.
• Any electrical conductor is an equipotential surface.

As a result, field lines always meet conductors at right angles (see the fourth field diagram in the last section).

Unlike with gravitational equipotentials there is quite a simple practical that the students can do to discover the shapes of electrical equipotentials in 2 dimensions.

Episode 406-2: Plotting equipotentials (Word, 69 KB)

Sum up by discussing the shapes of some common fields and their equipotential lines. Note that where the field is uniform, the equipotentials are evenly spaced, but in a non-uniform field equipotentials get further apart as the field decreases in strength (see episode 408).

(Note: the full lines represent the electric field and the dotted lines the equipotentials)

Student questions: Field lines and equipotentials

Episode 406-3: Fields lines and equipotentials (Word, 87 KB)

This episode introduces Coulomb’s law, which gives the force between two charges, in exactly the same way that Newton’s Law of Universal Gravitation gives the force between two masses. In fact, we will see that the two laws are identical in structure.

Lesson Summary

• Discussion: Coulomb’s law (15 minutes)
• Worked examples: Calculations involving Coulomb’s law (25 minutes)

Discussion: Coulomb’s law

We know that a field exists around a charge that exerts force on other charges placed there, but how can we calculate the force? The force will be dependent upon the sizes of the charges, and their separation. In fact the force follows an inverse square law, and is very similar in form to Newton’s Law of Universal Gravitation. It is known as Coulomb’s law, and it is expressed as:

F = k Q₁ Q₂r²

where Fis the force on each charge (N)

Q₁ and Q₂ are the interacting charges (C)

r is the separation of the charges (m)

The k is a constant of proportionality (like G in Newton’s Law of Universal Gravitation). In a vacuum, and to all intents and purposes, in air, we have

k = 9.0 × 10⁹ N m² C^-2 (units obtained by rearranging the original equation)

More traditionally, Coulomb’s law is written:

F = Q₁ Q₂4 π ε₀ r²

where ε₀ is known as the permittivity of free space; ε₀ = 8.85 × 10^-12 F m^-1 (farads per metre). Permittivity is a property of a material that is indicative of how well it supports an electric field, but is beyond the scope of these notes. Thus, we have k = 14 π ε₀. Different materials have different permittivities, and so the value of k in Coulomb’s law also changes for different materials.

Points to bring out about Coulomb’s law:

The form is exactly the same as Newton’s law of universal gravitation; in particular, it is an inverse-square law.

This force can be attractive or repulsive. The magnitude of the force can be calculated by this equation, and the direction should be obvious from the signs of the interacting charges. (Actually, if you include the signs of the charges in the equation, then whenever you get a negative answer for the force, there is an attraction, whereas a positive answer indicates repulsion).

Although the law is formulated for point charges, it works equally well for spherically symmetric charge distributions. In the case of a sphere of charge, calculations are done assuming all the charge is at the centre of the sphere.

In all realistic cases, the electric force between 2 charges objects absolutely dwarfs the gravitational force between them, as the first of the worked examples will show.

Worked examples: Calculations involving Coulomb’s law

Episode 407-1: Worked examples – Coulomb’s law (Word, 27 KB)

This episode introduces the above three quantities for the electric field. The students’ familiarity with the equivalent concepts in the gravitational field should help here. The major difference is that because of repulsion, by defining the zero of potential to be at infinity, we can have positive potentials and positive energies in the electric field, whereas they are always negative in the (solely attractive) gravitational field.

Lesson Summary

• Discussion: Field strength (5 minutes)
• Worked examples: Field strength (10 minutes)
• Discussion: Energy and potential (15 minutes)
• Demonstration: Potential around a charged sphere (30 minutes)
• Discussion: Field strength and potential gradient (10 minutes)
• Worked examples: The non-uniform electric field (25 minutes)

Discussion: Field strength

Recall: how is field strength defined at a point in the gravitational field? (As the force per unit mass placed at that point in the field – with units therefore of N kg^-1 .)

What would therefore be the natural way to extend this definition to the electric field? (As the force per unit charge. Thus it would have units of N C^-1.)

We thus define the electric field strength at a point in a field as:

E = FQ

where E is the electric field strength (N C^-1)

F is the force on charge Q at that point if the field

Important notes:

• The field strength is a property of the field and not the particular charge that is placed there. For example, at a point where the field strength is 2000 N C^-1 , a 1 C charge would feel a force of 2000 N whereas a 1 mC charge would feel a force of 2 N; the same field strength, but different forces due to different charges.
• The field strength is a vector quantity. By convention, it points in the direction that a positive charge placed at that point in the field would feel a force.

As will be explained in the next episode, the unit for electric field strength can also be expressed as volts per metre, V m^-1.

Now, for the non-uniform field due to a point (or spherical) charge, we can use Coulomb’s law to find an expression for the field strength. Consider the force felt by a charge q in the field of another charge Q, where the charges are separated by a distance r:

F = kQqr²

by Coulomb’s Law.

But E = Fq and so

E = kQr²

This is our result for the field strength at a distance r from a (point or spherical) charge Q .

Worked examples: Field strength

Episode 408-1: Field strength (Word, 31 KB)

Discussion: Energy and potential

We now turn to considerations of energy. Again, just as in the gravitational case, we choose to define the zero of energy at infinity. However, because of the existence of repulsion, we have the possibility of positive energy values as well as negative ones.

Consider bringing a positive charge q from infinity towards a fixed, positive charge Q . Because of the repulsion between the charges, we must do work on q to bring it closer to Q . This work is stored in the electric field around the charge q . The same would apply if both charges were negative, due to their mutual repulsion. In both cases therefore, the energy of q increases (from zero) as it approaches Q ; i.e. energy of charge q is positive.

If Q and q are of opposite signs, however, they attract each other, and now it would take work to separate them. This work is stored in the electric field around the charge q, and so q's energy increases (toward zero) as their separation increases; i.e. q's energy is negative.

With the aid of integration, we can use Coulomb’s law to find the energy of q in the field of Q . The final expression turns out as:

energy, E_E = kQqr

where r is the separation of the charges. E_E is measured in joules, J.

Notes on this expression:

By considering the work done if q were fixed and Q were brought up from infinity, it should be clear that this expression is the energy of either charge in the other’s field.

This expression is valid for the non-uniform field around point or spherical charges.

By defining energy to be zero at infinity, this expression gives us absolute values of energies. Differences can we worked out as change in:

Δ E_E = kQq(1r₁  −  1r₂).

It is essential to include the sign of the charges in this expression to get the correct E_E. Thus two positive or two negative charges yield positive E_E values, whereas opposite charges yield negative E_E values, as we would expect from our discussion above.

We can now go on to discuss potential. In the gravitational field, potential at a point was defined as the energy per unit mass at that point. The natural extension to electric fields is therefore as the energy per unit charge:

V = E_EQ

V is therefore measured in joules per coulomb, J C^-1, which has the alternative, and more familiar name of volts, V. (Students sometimes worry about the fact the potential has the symbol V and its unit is also V for volts – the context makes it clear which V is being used, but be aware of this possibility for confusion).

Note again that potential is a property of the field and not the individual charge placed there. Thus at a point in a field where the potential is 500 V, a 1 mC charge has an energy of  0.5 J whereas a 1 C charge has an energy of 500 J. (One might say that potential is to energy exactly as field strength is to force in the field).

For the non-uniform field around a point charge, the above expression for E_E gives us a simple expression for potential. The potential at a distance r from a point or spherical charge Q is given by:

Note that this gives positive or negative values for potential depending upon whether Q is positive or negative. (Remember though that a negative charge placed where there is a negative potential will have a positive energy as expected due to the repulsion of the two negative charges).

For the non-uniform field around a point charge, the above expression for E_E gives us a simple expression for potential. The potential at a distance r from a point or spherical charge Q is given by:

V = kQr

Note that this gives positive or negative values for potential depending upon whether Q is positive or negative. (Remember though that a negative charge placed where there is a negative potential will have a positive potential energy as expected due to the repulsion of the two negative charges).

Demonstration: Potential around a charged sphere

Exploring the electrical potential near a charged sphere. This makes use of a flame probe; it is essential to practise the use of this before demonstrating in front of a class.

Episode 408-2: Potential near a charged sphere (Word, 102 KB)

Episode 409-2: Flame probe construction (Word, 237 KB)

Discussion: Field strength and potential gradient

As with the gravitational field there is a deep and important connection between the rate at which potential changes in a field, and the field strength there. Quite simply, the field strength is equal to the negative of the rate of change of potential:

E = −  Δ E_Echange in distance

The minus sign is again a relic of the more precise vector equation – it is there to give the direction of the field strength. The diagram below shows a plot of potential around a positive charge, and the two gradients drawn show where the field is high and where it is low. (Note that both of these gradients are negative. This would give a positive field strength by

E = −  Δ V Δ r , indicating that the field acts in the positive x-direction, as indeed it does. This is what was meant by the minus sign being a relic of the more precise vector equation).

We can now also understand why it is that a conductor is an equipotential surface. Inside a conductor, no electric field can exist – if it did, the charges would feel a force and move around in such a way as to reduce the field. At some point equilibrium is reached and the field is zero. If the field is zero, then the potential gradient must be zero – i.e. the conductor is an equipotential surface.

It should now also be clear why equipotential surfaces get further apart as the field decreases in strength as was seen in the diagrams in episode 1, and why field strength can be quoted in units of V m^-1.

Worked examples: The non-uniform electric field

Episode 408-3: Non-uniform electric fields (Word, 34 KB)

So far we have mainly concentrated on the non-uniform fields around point or spherical charges. We will now discuss the physics of the uniform electric field, such as that between 2 parallel charged plates. The basic definitions of field strength and potential lead to different results than those for the non-uniform field.

Lesson Summary

• Discussion: Uniform electric fields (5 minutes)
• Demonstration: Potential and field strength in a uniform field (25 minutes)
• Discussion: Accelerating charges through a potential difference (10 minutes)
• Student questions: Uniform electric fields (10 minutes)
• Student questions: Millikan’s oil drop experiment (10 minutes)
• Discussion: Comparison of gravitational and electric fields (10 minutes)

Discussion: Uniform electric fields

What do we mean by a uniform field? (One that does not vary from place to place. In terms of the field lines, this means that they are parallel and evenly spaced. Also as field strength =   −  (potential gradient), the equipotentials should also be evenly spaced.)

Where have we seen such as field? (The field between two parallel charged plates.)

How will a charge move in such a field? The force is given by

F = E × Q

Since E is constant, the force will be constant and therefore, by

F = m × a

the acceleration will also be constant and directed along the field lines for a positive charge and opposite to the field lines for a negative charge. (Note: A constant acceleration produces a parabolic path for the charge if it has any component of motion across the field lines, just as with projectiles in the Earth’s constant gravitational field near its surface).

Demonstration: Potential and field strength in a uniform field

A flame probe can be used to test the potential in the uniform field. Also, plotting a graph of potential against distance between the plates gives us the field strength because

E = – (potential gradient)

This demo will confirm the fact that the equipotentials are evenly spaced and parallel to the plates, giving a uniform electric field perpendicular to the plates.

Episode 409-1: Potential and field strength in a uniform electric field (Word, 65 KB)

Episode 409-2: Flame probe construction (Word, 237 KB)

Now, since E = – (potential gradient), but the potential gradient is uniform as we have seen, the electric field strength will simply be given (in magnitude) by:

E = Vd

where:

• V is the potential difference between plates (in volts)
• d is the distance between plates (in metres)

The direction of the field is obviously from the positive to the negative plate.

Discussion: Accelerating charges through a potential difference

What happens when a charge is accelerated through a potential difference? If there is a potential difference between two points, there must be a field between them since E = – (potential gradient). Thus a charge that moves between the two points under the action of the field has work done on it by the field (as work = force × distance and a force is provided by the field). For a potential difference of V between two points, the work done by the field on a charge Q as it moves between those points is given by: W = V × Q

where W is work done in joules, J.

(Careful here – so far we have used V as the symbol for potential, and V as the unit for potential, volts. Now we are using it for a potential difference . If this is likely to be a source of confusion, you may wish to adopt an expression like

W = Q ×  Δ V , with Δ meaning change in).

We can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it.

(Note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 V,

i.e. W = 1 eV

W = 1.6 × 10^-19 C × 1 V

W = 1.6 × 10^-19 J).

In fact this expression is really nothing more than the statement that potential is the energy per unit charge. In this case, V = WQ is simply stating that the change in energy (or work done) per unit charge is equal to the change in potential (or potential difference), V .

Student questions: Uniform electric fields

Episode 409-3: Uniform electric fields (Word, 43 KB)

Worked examples: Millikan’s oil drop experiment

Episode 409-4: Millikan’s oil drop experiment (Word, 42 KB)

Discussion: Comparison of gravitational and electric fields

We finish this topic by discussing the parallels between gravitational and electric fields. You could ask your students to show the parallels in their own way – a useful summarising activity. Here are the main points that they should come up with:

We have seen the following similarities:

• Both fields follow an inverse square law for both force and therefore field strength
• For both types of field, potential and energy stored in the field are inversely proportional to distance
• We define the zero of potential to be at infinity
• For attractions (and therefore always for gravitational fields) energies are always negative. As work always has to be done on the object (charge or mass) to keep it at that point in the field.
• The major difference is that repulsions occur in electrostatic fields between charges of the same sign, whereas as far as we know, gravity is always attractive

The similarity of the fields may be brought home further by comparing the forms of the equations.