Fields Tutorial 4  Electric Fields
Coulomb’s
Law
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") 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} Nm^{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]
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
(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. 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.
Question 1 
Describe the similarities and differences between the forces in a gravity field and electric field. 

Question 2 
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? 
Electric Forces as Vectors
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}
2. Like charges in line
Now, let’s make all the charges positive:
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.
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 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? 
(a) Work out the forces F_{1} and F_{2}:
F_{1} = 9 × 10^{9} × 15 × 10^{9} × 2 × 10^{9} = 12 N (150 × 10^{6})^{2}
Similarly: F_{2} = 9 × 10^{9} × 34 × 10^{9} × 2 × 10^{9} = 42.5 N (120 × 10^{6})^{2}
F = 42.5 N – 12 N = 30.5 N to the left.
(b) F_{1} = F_{2 }
Therefore:
9 × 10^{9} × 15 × 10^{9} × 2 × 10^{9} = 9 × 10^{9} × 34 × 10^{9} × 2 × 10^{9} d_{1}^{2} d_{2}^{2}
Cancelling:
d_{1}^{2} d_{2}^{2}
d_{2}^{2} ÷ d_{1}^{2} = 34 ÷ 15 = 2.27
d_{2} = √2.27 × d_{1} = 1.51 × d_{1}
We also know that d_{1} + d_{2} = 150 + 120 = 270 mm
Thus: d_{1} = (270 – d_{2})
Therefore
d_{2} = 1.51 × (270 – d_{2})
d_{2} + 1.51 d_{2} = 406
d_{2} = 406 ÷ 2.51 = 162 mm
Therefore: d_{1} = 270 – 162 = 108 mm

Electric
Field Strength
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 = F/Q
[E
– electric field strength;
F – force;
Q – charge]
E is a vector quantity and its units are newtons per coulomb, N/C. It can be negative (attractive) or positive (repulsive).

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, we can say that:
F = F_{1} + F_{2} = qE_{1} + qE_{2}
E = F = qE_{1} + qE_{2}
q q
2. Like charges in line
We can use a similar argument to the one above:
E = F = qE_{1}  qE_{2}
q q
If all the charges were negative, the same would apply.
3. Charges not in line
E^{2} = E_{1}^{2} + E_{2}^{2}
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.
Uniform Electric Fields
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 = V/d
[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.
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? 
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 AS Physics 2. 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.
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 an 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. 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. 
Horizontal velocity remains unchanged at 3.7 ´ 10^{7} m/s, since there is no horizontal force. 
We can now
work out the upwards force using
F = QE
F
= 1.6
´
10^{19}
´
23000 = 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} ms^{2}. Upwards velocity = at = 4.04 ´ 10^{15} ms^{2} ´ 1.08 ´ 10^{9} s = 4.37 ´ 10^{6} m/s

Now we can do the vector addition to work out the resultant velocity
v^{2}
= (3.7
´
10^{7} m/s)^{2} + (4.37
´
10^{6} m/s)^{2} = 1.388
´
10^{15} m^{2}s^{2}
Þ
v = 3.73
´
10^{7} m/s 
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 3.73
´
10^{7} m/s 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.
An alpha particle enters a
uniform electric field as shown:
The plates are 5 cm long.
