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 1 
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? 
We
can define the electrical potential as the energy
per unit charge. It has the code
V_{e}
and is given by the equation:
The units are Joules per Coulomb (JC1) 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.
Suppose we have the positive plate in a uniform field at +300 V, and the negative plate is at 0 V. Suppose we have 10 equipotentials equally spaced in the uniform field. Each equipotential has a potential difference of + 30 V. So there are equipotentials at 0 V, 30 V, 60 V, 90 V, etc.
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 V_{e} = __1__ Q = 9 × 10^{9} × 1.6 × 10^{19 }= 28.8 V 4pe_{0} 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 × 10^{11}
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 Xrays). 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 mv^{2}
Ž v^{2} = 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 the equation
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 4 
How would the trajectory of a
proton be different? 

Question 5 
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 ´ 10^{7} m/s just as they enter the electric field.
(Electronic
charge = 1.6 ´
10^{19}C; 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? 
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__
4pe_{0}
where
e_{0}
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
r^{2}. 
Inversely
proportional to
r^{2}. 
Force
Equation 
F = __1__ Q1Q2 4pe_{0} r^{2} 
F = G m_{1}m_{2} r^{2 } 
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.