Fields Tutorial 4  Electric Fields
This is quite a long tutorial. You may want to take a break half way through.
Electric fields arise due to static electricity. Static electricity arises because charges are separated. Like charges repel, unlike charges attract. Positive charges result from too few electrons, negative from too many electrons. Electric fields are areas around a charge in which a force is felt. This force can be attractive or repulsive. Gravity fields are only ever attractive.
In the diagram above, we have charged up a metal can with positive charge by
removing electrons. We have placed it on a square of insulating material.
When we bring a metal sphere that is connected to the ground (earthed),
electrons are attracted to the positive charges. They form an electric
field. If the electric field is strong enough, the insulating properties
of air break down and a spark jumps. The charge on the can becomes neutral
as the positive charges are balanced by the negative charges.
Positive charges NEVER move in solid materials. 
There
are many parallels between gravity fields and electric
fields. The electric field is
the region of force around a point charge.
In 1784 Charles Augustine de Coulomb (pronounced "Collume"
not "Coolomb") showed that the force between two point charges was
proportional to each charge and inversely proportional to the square of the
distance between them. This reminds
us of Newton’s Law of Gravitation.
Coulomb’s
Law can be summarised in code as:
We
can rewrite this proportionality as:
The
constant k
has a value of 8.99
´
10^{9} N m^{2} C^{2} for two charges in a vacuum.
The constant is not normally written as
k,
but instead:
[e_{0}
= 8.85
´
10^{12} C^{2} N^{1} m^{2} or F m^{1}]
This
term
e_{0}
(‘epsilon nought’ 
e
is a Greek letter ‘e’) is called the permittivity
of free space. We can write the
units for
k
as m F^{1}
(metres per Farad).
The equation is written as:
A useful dodge is:
We
can approximate the value of
k to 9.0
´
10^{9} m F^{1}. Its large value
tells us why electrical forces are very strong compared to gravity forces, even
when the charges are quite small. This
helps us to explain why solid materials can exist at all.
The force F is positive when the charges are both positive or both negative; this indicates that the force is repulsive, which is borne out by our elementary observations of electrostatics. If one charge is positive and the other is negative, we can easily see that the force F is negative, therefore attractive.
The graph below shows the variation of force with distance.
The data used to generate this graph used two like charges each of 5.0 nC, while the distance was between 1.0 × 10^{3} m and 5.0 × 10^{3} m
Describe the similarities and differences between the forces in a gravity field and electric field. 

Find the electrostatic force between a proton and an electron in a hydrogen atom if their separation is 5.3 ´ 10^{11} m. What does the sign tell you? 
The simplest way of showing how electric fields work is with field lines. In the diagram below we see the field lines coming from a negative point charge and a positive point charge.
The field lines go from positive to negative. So for the negative charge they point inwards. For the positive charge they go outwards. The field of a point charge is radial.
Unlike gravity fields (which are always attractive), electric fields can be attractive or repulsive. Let's look a a field between two unlike charges of the same value:
Notice how the field lines run from the positive to the negative charge. At point X, the lines are parallel. Since the charges are of the same magnitude, the point X is half way between the two.
Now let's see what happens when the two charges are both positive charges:
At point X there is a neutral point where the forces add up to zero. The charges are of the same magnitude, so the neutral point is half way between them. We see the same for two negative charges of the same magnitude:
We can see that the neutral point is half way between. If the magnitudes were different, the neutral point would lie at a point that can be worked out using the square root of the ratio of the magnitudes of the charges.
The electric field between two parallel plates is uniform.
In a uniform field, the field lines are parallel. However the lines at the end bulge out slightly. We ignore this effect.
All the diagrams show the fields in two dimensions. In reality they are threedimensional
While gravity fields and forces are attractive, electric forces and fields can be repulsive or attractive. Force and field are both vector quantities in both situations. Therefore, if we place a point charge between two other charges we can consider three different situations.
1. Unlike charges in line
In this first situation, the test charge q experiences a force, F_{1} which is attractive because the charge Q_{1} is opposite. At the same time, it also experiences a repulsive force, F_{2} because Q_{2} has the same sign. Therefore:
F_{res} = F_{1} + F_{2}
The test charge accelerates towards the charge Q_{1}.
2. Like charges in line
Now, let’s make all the charges positive:
The test charge +q is repelled by +Q_{2} with a force F_{2}. It is also repelled with a force F_{1} by Q_{1}.
This time the forces are in opposite directions. Therefore:
F_{res} = F_{1} – F_{2}
There will be a point at which the forces are equal and opposite, so the test charge stays where it is. This is when forces F_{1} and F_{2} will balance so that F_{res} = 0.
3. Charges not in line
The third situation is when the test charge is NOT in line with the other two charges:
In this case, the resultant force is the vector sum of the two forces, i.e:
(F_{res})^{2} = F_{1}^{2} + F_{2}^{2}
We can apply the rules for any number of charges
around a test charge.
Worked example A 15 nC charge is placed 150 mm from a test charge of +2.0 nC. The third charge of 34 nC is placed 120 mm from the test charge, as shown below:
(a) What is the resultant force on the test charge? (b) At what distance will the resultant force be zero? 
Answer (a) Work out the forces F_{1} and F_{2}:
F_{1} = 9.0 × 10^{9} F m^{1} × 15 × 10^{9} C × 2.0 × 10^{9} C = 12.0 N (150 × 10^{6} m)^{2}
Similarly: F_{2} = 9.0 × 10^{9} F m^{1} × 34 × 10^{9} C × 2.0 × 10^{9} C = 42.5 N (120 × 10^{6} m)^{2}
F = 42.5 N – 12.0 N = 30.5 N to the left.
(b) F_{1} = F_{2 }
Therefore:
9.0 × 10^{9} m F^{1} × 15 × 10^{9} C × 2.0 × 10^{9 } C = 9 × 10^{9} m F^{1} × 34 × 10^{9} C × 2.0 × 10^{9} C d_{1}^{2} d_{2}^{2}
Cancelling:
d_{1}^{2} d_{2}^{2}
d_{2}^{2} ÷ d_{1}^{2} = 34 C ÷ 15 C = 2.27
d_{2} = √(2.27) × d_{1} = 1.51 × d_{1}
We also know that: d_{1} + d_{2} = 150 mm + 120 mm = 270 mm
Thus: d_{1} = (270 mm – d_{2})
Therefore
d_{2} = 1.51 × (270 mm – d_{2})
d_{2} + 1.51 d_{2} = 406 mm
d_{2} = 406 mm ÷ 2.51 = 162 mm
Therefore: d_{1} = 270 mm – 162 mm = 108 mm

A
gravity field is where a force is exerted on a mass; an electric field is a
region where a force is exerted on a charge.
The electric field strength is defined as:
force per unit charge.
In
equation form this is represented as:
[E
– electric field strength (N C^{1});
F – force
(N);
Q – charge
(C).]
E is a vector quantity and its units are newtons per coulomb, N C^{1}. It can be negative (attractive) or positive (repulsive).
What
does it mean when the force per unit charge is
(b)
Positive 
Since
the electric field strength is defined as force per unit charge, we can write:
This
equation is true for all point charges, which can be considered to have a radial field. A spherical
charge can also be considered to have a radial field provided the distance from
the charge to the centre is much greater than the radius of the sphere.

A charge of +1.6 ´ 10^{19} C has a force of 8.7 ´ 10^{15 }N exerted on it when it is placed a certain point in a radial electric field. What is the electric field strength? 
Electric Field Strength as a Vector
While gravity fields and forces are attractive, electric forces and fields can be repulsive or attractive. Force and field are both vector quantities in both situations. Therefore, if we place a point charge between two other charges we can consider three different situations.
1. Unlike charges in line
Since E = F/q, and F = Eq, we can say that:
F = F_{1} + F_{2} = qE_{1} + qE_{2}
Therefore, with the cancellation of the q terms:
E_{res} = E_{1} + E_{2}
2. Like charges in line
We can use a similar argument to the one above:
Therefore, with the cancellation of the q terms:
E_{res} = E_{1}  E_{2}
If all the charges were negative, the same would apply.
3. Charges not in line
In this case, the resultant force is the vector sum of the two fields, i.e:
We can apply the rules for any number of charges
around a test charge.
We can plot a graph of electric field strength against distance:
We
can see that the graph shows a classic inverse square law.
Two point charges are placed at a distance of 8.0 ´ 10^{3} m apart as shown in the diagram.
(b) What is the force acting between the two points? What can you say about the force? (c) What is the value of the electric field strength at the midpoint between the charges? 
The direction of the field is the direction of the force.
The force in this case is the force exerted by the electric field on a
hydrogen ion (proton).
If we are looking at the electric field between two point charges, we need to take both into account both the charges.
Where two charged plates are close together, the radial fields of the charges combine to make a uniform field.
Notice that the field bulges at the ends; generally we ignore this.
In
this case we can show that the electric field strength is given by a simpler
relationship:
[E
– electric field strength;
V –
voltage; d distance in m]
Strange as it may appear, the units of volts per metre are entirely consistent with newtons per coulomb.
The relationship between the potential difference and the distance shows direct proportionality (i.e. a straight line of positive gradient that goes through the origin. You remembered that, didn't you?).
Two
plates are placed 15 cm apart connected to a 4500 V supply.
(a)
What is the strength of the uniform field between the plates? (b) What is the force acting on a 6 nC test charge between the plates? 
Breakdown Voltage in Insulators
Insulating materials are by their nature poor conductors of electricity. They are vital for the correct functioning of electric circuits. However there is a limit to their insulating properties. Electrical engineers often refer to the breakdown voltage of the insulator. For any insulating material, the breakdown voltage depends on the thickness of the insulating layer. A layer of PVC that is 1 mm thick will safely insulate 230 V, but will not insulate 230000 V.
The breakdown voltage is actually the electric field strength. It is also called the dielectric strength of a material. The table below shows a few breakdown voltages of materials.
Material 
State 
Breakdown voltage / V m^{1} 
Air 
Gas 
3.0 × 10^{6} 
Distilled water 
Liquid 
70 × 10^{6} 
Polythene 
Solid 
18 × 10^{6} 
Diamond 
Solid 
2000 × 10^{6} 
In data books, the breakdown voltage is often given in kV mm^{1}.
The Cathode Ray Tube
We study the trajectory of charged particles in electric fields using apparatus like this. It called a cathode ray tube.
(It is a very expensive piece of kit, so handle it with care. Always wear goggles when using it, as it's under high vacuum and could implode.)
The cathode ray tube contains an electron gun that produces a stream of electrons. The tube is evacuated so that the chances of a collision with a gas molecule are much reduced.
The electron gun works like this:
Electrons are boiled off a hot cathode heated up by a filament (just like a light bulb). This is called thermionic emission.
The electrons are accelerated by being attracted towards the anode by the electric field set up between the cathode and anode.
Most electrons collide with the anode.
A few pass through a hole in the middle of the anode.
These form a fine beam.
The fine beam strikes a phosphor screen to show up as a bright green dot.
If there is a potential difference across the plates, a second electric field is set up to deflect the electron beam.
We can derive an equation for the energy of the accelerated electrons. The kinetic energy of the electrons is given by:
E_{k} = Particle kinetic energy (J)
e = charge on the particle (1 e = 1.6 × 10^{19} C )
V = accelerating voltage (V)
We also know the simple equation for kinetic energy in terms of mass and speed:
v = speed of particle (m s^{1})
m = mass of the particle (kg)
So we can combine these two to give:
Rearranging:
To give our final result:
This is the sort of equation you may be asked to prove in the exam.
Trajectory of charged particles in a uniform electric field
Charged particle such as electrons behave in exactly the same way as projectiles do in a gravity field. You may want to revise the ideas from Mechanics Tutorial 9. You will remember that:
The vertical and horizontal movements are independent.
The horizontal velocity stays the same.
The vertical velocity changes because of the acceleration due to the gravity field.
Exactly the same applies with electric fields.
The electron is attracted by the positive plate:
Its horizontal velocity is not changed at all.
Its vertical velocity starts at zero as it enters the field, and increases to v_{y} (in an upwards direction) as it leaves the field.
The path is a parabola.
The path of the electron is only a parabola within the electric field. Once the electron has left the electric field, its path is a straight line. 
Unlike a stone that is thrown horizontally that will hit the ground, once an electron leaves the electric field, it will continue in a straight line. It will have a resultant velocity that is the vector sum of the vertical and horizontal velocities.
Use the equations of motion to analyse the two different movements. Let us look at the trajectory of an electron using the following principles:
The electric field goes from positive to negative.
The electron carries a charge of –1.6 × 10^{19} C.
The electron is attracted to the positive side, so goes against the field.
The proton carries a charge of +1.6 × 10^{19} C.
The proton is repelled by the positive side, so moves in the direction of the field.
Other charged particles have whole number
multiples of 1.6 × 10^{19} C.
Deriving an expression for the trajectory
Consider an electron of charge
e
C and mass
m
kg travelling at a speed of
v
m s^{1} into the space between two plates that are
l m long and
d m apart.
The
plates have an electric field of
E
V m^{1} between them.
We know that electric field strength in a uniform field is given by:
We know by definition that:
F = EQ
So we can combine the two equations to give:
The application of Newton II gives us:
Therefore:
The resulting velocity, v_{res}, consists of a horizontal component, v_{x}, and a vertical component v_{y}:
Horizontal velocity remains the same, so we can work out the time using:
To work out the vertical velocity, use:
We can write an expression that will give us the vertical velocity by combining all of the expressions above:
This is another expression you might well be asked to derive for an exam question. This expression is quite useful in an exam, because we are likely to be given:
the speed of the electron;
the voltage across the plates;
the distance between the plates;
the length of the plates.
We know from databooks that the mass of an electron is 9.11 × 10^{31} kg and the charge is 1.60 × 10^{19} C. If you can't remember the relationship, go back to first principles. You will get the full credit for a correct answer.
We work out the resultant speed of the electron by Pythagoras:
The angle is worked out using:
This worked example uses first principles.
Worked Example
The
electric field strength between a pair of plates length 4.0 cm in a
cathode ray tube has a value 23 000 N C^{1}. An electron enters the
field at right angles at a velocity of 3.7 ×
10^{7} m/s from left to right. What is the velocity and direction of
the electron as it leaves the field?

Before we attempt our answer, we need to think about how the electron is going to travel. It is travelling in a straight line as it enters the electric field. As it goes in, there will be a force acting on the electron that attracts it to the negative plate. This will cause the electron to accelerate towards the positive plate. Since its horizontal velocity is unchanged and its vertical velocity is increasing, it describes a parabolic path. (If the plates were long enough, or if the electron were slow enough, the electron would eventually hit the positive plate.) The electron leaves the field and travels again in a straight line, but this time at an angle to its undeviated path. We can work out the resultant velocity by consideration of the horizontal and vertical velocities. 
Answer Horizontal velocity remains unchanged at 3.7 ´ 10^{7} m s^{1}, since there is no horizontal force. 
We can now
work out the upwards force using
F = QE
F
= 1.6
´
10^{19} C
´
23000 N C^{1} = 3.69
´
10^{15} N
The minus sign tells us that the force is attractive, and is against the direction of the field, which is from positive to negative. However we will ignore the minus sign from now on.

We
now need to work out the upwards velocity of the electron, which we do by
multiplying the acceleration by the time.
We need to know the acceleration and the time.
Simple enough, really.
Time
taken to travel through the field = distance
¸
horizontal speed
= 0.04 m
¸
3.7
´
10^{7} m s
= 1.08
´
10^{9} s
Then
we use Newton II to find acceleration,
a = F/m.
a
= 3.69
´ 10^{15} N
¸ 9.1
´
10^{31 }kg = 4.04
´
10^{15} m s^{2}.
Upwards velocity = at = 4.04 ´ 10^{15} m s^{2} ´ 1.08 ´ 10^{9} s = 4.37 ´ 10^{6} m s^{2}

Now we can do the vector addition to work out the resultant velocity.
v^{2}
= (3.7
´
10^{7} m s^{1})^{2} + (4.37
´
10^{6} m s^{1})^{2} = 1.388
´
10^{15} m^{2 }s^{2}
Þ
v = 3.73
´
10^{7} m s^{1} 
Now
we can work out the angle of deflection,
q.
q
= tan^{1} 4.37 = tan^{1} 0.118 = 6.7 ^{o}
37
The
velocity is 3.7
´
10^{7} m s^{1} in a direction of 6.7 ^{o} to the horizontal.

The charged particle can, of course, be positive. A proton can pass through an electric field, as can an alpha particle, or a positive metal ion. Protons are made by ionising hydrogen gas. Alpha particles are obtained from an alpha source. Such apparatus would be available in a university physics department, but not in a school or college lab.
An alpha particle enters a uniform electric field as shown:
The plates are 5.0 cm long.
