Topic 9 -
Electric Fields
| Key Words Electric Field, charge, permittivity of free space, attractive, repulsive |
|
You are expected to · be able to recall and use the equation:
· be familiar with the concepts of the electric field and to be able to use the formulae E = F/Q, E = V/d, and
· be able to use the formula
·
be able to describe how charged particles are affected by electric
fields. · able to understand the similarities and differences between electric and magnetic 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 Coulomb 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 ´
109 Nm2 C-2 for two charges in a vacuum.
The constant is not normally written as k,
but instead:

[e0
= 8.85 ´
10-12 C2 N-1 m-2 or F/m]
This
term e0
(‘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.
The equation is written as:

Since you have to know this formula, the easiest way is:

We
can approximate the value of k to 9.0 ´
109 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. |
ANSWER |
| 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? |
ANSWER |
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.
|
||||||||
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.

|
|
We can plot a graph of electric field strength against distance:

We
can see that the graph shows a classic inverse square law.
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.
| 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? |
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.
|
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? |
Let
us look at a more complicated 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 ×
107 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 un-deviated path. We can work out the resultant velocity by consideration of the horizontal and vertical velocities. |
|
Horizontal velocity remains unchanged at 3.7 ´ 107 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 ´
107 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 ´
1015 ms-2. Upwards velocity = at = 4.04 ´ 1015 ms-2 ´ 1.08 ´ 10-9 s = 4.37 ´ 106 m/s
|
|
Now we can do the vector addition to work out the resultant velocity v2
= (3.7 ´
107 m/s)2 + (4.37 ´
106 m/s)2 = 1.388 ´
1015 m2s-2 Þ
v = 3.73
´
107 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 ´
107 m/s in a direction of 6.7 o to the horizontal.
|
|
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? |
When
we looked at gravity fields, we saw that gravitational potential energy is the
energy needed to bring an object from infinity to a certain point.
We also saw that gravitational potential was the energy per unit mass.
There are two similar quantities in electric fields, potential
energy and electric potential.
Suppose we moved a charge form one point to another in an electric field. We would have to do a job of work against the field (or get work out if it were with the field), so there is an energy change. We can show this as a graph:

The
area under the graph represents the energy
transformed. Using the
mathematical trick of calculus, we can show that the energy is given by the
equation:

| Question 8 |
A positive charge of 6 nC attracts a negative charge of 5 nC from a distance of 4 mm. What is the work done? What does the sign mean? |
ANSWER |
We
can define the electrical potential as the energy
per unit charge. It has the code Ve
and is given by the equation:

The units are Joules per Coulomb (JC-1) or Volts (V). Indeed the volt is defined as the energy transformed when unit charge is moved between two points. We can show the relationship between the electric field and potential in this graph. Potential is the area under the graph:

All points within an electric field that have the same potential are called equipotentials, rather like contours on a map. In a radial field, the equipotentials are concentric circles. In a uniform field, they are parallel lines equally spaced.

| Why are equipotentials not equally spaced in a radial field? |
Worked Example
| If a hydrogen atom consists of a proton and an electron at an average separation of 5.0 × 10-11 m, what is the electric potential at this distance and the potential energy of the electron? What would the field strength be? |
|
Use Ve = __1__ Q = 9 × 109 × 1.6 × 10-19 = 28.8 V 4pe0 r 5.0 × 10-11 |
| Potential energy = charge × potential = 28.8 V × 1.6 ×10-19 C = 4.6 × 10-18 J |
|
Field
Strength = potential ÷ separation =
28.8 ÷ 5.0 × 10-11 = 5.76 × 1011
N/C |
|
A tiny negatively charged oil drop is held stationary in an electric field between two horizontal parallel plated as shown. The mass is 2.5 ´ 10-10 kg.
(a)
What are the two forces acting on the drop and in which direction do they
act? (b)
The drop is stationary. What
can be said about the two forces? (c) What is the charge on the oil drop? |
The
key points to remember are:
·
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.
Electrons can be accelerated by an electric field. This happens in any cathode ray tube, e.g. a TV set, or a VDU. Electrons are “boiled off” a red hot cathode in a process called thermionic emission. They are then attracted by a positive electrode, the anode, at a high voltage. This makes them move forward at a high speed. The electrons move parallel to the direction of the field. Most hit the anode, but some fly through a hole at the front, to hit a screen at the far end of the tube. The screen is covered in phosphor, and the energy the electrons have is converted to light (and some X-rays). Because this arrangement spits out electrons, it is called an electron gun.

The
electrons are repelled by the cathode and accelerated by the anode. Each
electron is given energy, which can be found using energy = charge × voltage, E
= QV. The energy is entirely kinetic
energy, so we can say that:
QV = Ek = ½ mv2
Þ v2 = 2QV
m
There is no reason why positive charges cannot be accelerated in the same way. Indeed in a mass spectrometer, positive ions are accelerated by an electric field and bent by a magnetic field to hit a detector. Particle accelerators work in the same way.
Let us consider how an electron behaves as it enters a uniform electric field at right angles:

We
know that the electric field strength is given by E = V/d
From our definition of electric field we can say that the force on the electron is given by F = EQ.
Therefore
the force is given by combining these two equations:

We
can apply Newton II to help us to work out the acceleration.
If we wanted to know the upwards velocity, we would work out the time
interval, and then use a = Dv/Dt
to calculate the upwards velocity.
Since the electrons are in a vacuum, the horizontal velocity is not
affected.
Once
we know the upwards velocity, then we can do a vector
addition to tell us the resultant velocity.
| Question 11 |
How would the trajectory of a
proton be different? |
ANSWER |
| Question 12 |
Two parallel plates are set at a distance of 12 mm apart in a vacuum. The plates are 30 mm long. The top plate has a potential of +300 V and the bottom has a potential of – 300 V as shown. The electrons have been accelerated to a horizontal velocity of 2 ´ 107 m/s just as they enter the electric field.
(Electronic
charge = 1.6 ´
10-19C; mass of an electron = 9.11 ´
10-31 kg) (a)
Sketch on a copy of the diagram the electric field between the plates (b)
What is the electric field strength at a point midway between the plates? (c)
What is the force on an electron at this point? (d)
What is the resulting acceleration? (e)
What is the vertical velocity as the electron leaves the plates? (f) What is the resultant velocity? |
ANSWER |
Comparing
Electric and Gravitational Fields
There
are many analogies that can be drawn between electric fields and gravity fields.
Theoretical physicists would go as far as saying that the two are
possibly different manifestations of the same thing.
Let us compare the two:
|
Feature |
Electric
Field |
Gravity
Field |
|
Quantity
susceptible to the force |
Charge |
Mass |
|
Constant
of Proportionality |
__1__ 4pe0
where
e0
is the permittivity of free space. The
value of e can
be changed by adding a material. |
G The
value of G, the universal gravity
constant has the same value for all media, including a vacuum.
|
|
Relationship
with distance |
Inversely
proportional to r2. |
Inversely
proportional to r2. |
|
Force
Equation |
F
= __1__ Q1Q2
4pe0
r2 |
F = -G m1m2
r2 |
|
Direction
of force |
Can
be attractive or repulsive |
Always
attractive |
|
Relative
strength |
Strong
at close range |
Weak.
Can only be felt with massive objects |
|
Range |
Infinite |
Infinite |
The
gravitational attraction between particles in an atom is so small as to be
negligible. The nucleus and its
electrons are held together entirely by electrostatic forces, and these are
involved in chemical reactions.
Gravity
forces hold planets together and hold them in their orbits. Electrostatic forces over the interplanetary distances can be
ignored.
|
Summary Force between charges:
k =
1 = 9 ´ 109 m/F
4pe0 Electric Field Strength: Force per unit charge
E = F/Q For a uniform field:
E
= V/d Electric Potential: Energy per unit charge
Potential Energy: Energy needed to bring a charge from infinity to a second charge.
Ep = k Q1Q2 r |
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