Motion of Charges in
Electric Fields
Electric Potential Difference
• When a charged particle is placed in an electric field it
experiences a force F because of its charge.
• If we move the charged particle some distance s in the
direction opposite to that of the force we have to do work W
on the particle, as W = Fs.
• We define the electric potential difference ΔV between two
points in an electric field as the work W done per unit charge
q on a small positive test charge when it is moved between
the two points, provided that all other charges remain
undisturbed.
• ∆𝑉 =
𝑊
𝑞
• This formula gives the unit of potential difference as joules per
coulomb (JC-1). This is call the volt (V), where 1V = 1JC-1.
Electric Potential Energy
• When we move a charge in an electric field and do work on it
we give it energy, as work is the transfer of energy.
• We are actually increasing the electric potential energy of the
charge, similarly to the way we increase the gravitational
potential energy of a body when we lift it in Earth’s
gravitational field.
• The electric potential energy of a charge is the energy that the
charge has due to its location in an electric field
• By the conservation of energy the change in the potential
energy of the charge must be equal to the work done on it.
• Thus the formula can be written as:
• ∆𝑉 =
∆𝑃𝐸
𝑞
The electron-Volt (eV)
• One electron volt is the change in energy of an electron when it
moves through a potential difference of one volt.
• ∆𝑃𝐸 = 𝑞∆𝑉
•
= 1.6 x 10-19 x 1
•
= 1.6 x 10-19 J
• 1eV = 1.6 x 10-19 J
• Note:
• If an electron passes through a potential difference of 2000V, its energy
changes by 2000eV
• If a proton passes through a potential difference of 500V, its energy
changes by 500eV
• If an alpha particle passes through a potential difference of 3000V, its
energy changes by 6000eV
• To convert Joules to electron volts, divide by 1.6x10-19
• To convert electronvolts to Joules, multiply by 1.6x10-19
Example Question 1
• A charge of 5mC experiences a force of 0.2N in an electric
field. It is moved a distance of 20cm against this force. Find
the potential difference through which the charge has moved.
Example Question 2
• An electron is accelerated from rest through a potential
difference of 2000V. Find its kinetic energy in electron-volts
and in Joules.
Uniform Electric Fields
• Consider the uniform field between two parallel plates, d
metres apart, with a potential difference of ΔV volts between
them.
• Consider moving a small positive charge q from the negative
plate to the positive plate.
• Because it is a uniform field of constant strength E there is a
constant force (F = qE) on this charge throughout its travel. To
move this charge we have to do work on it.
Uniform Electric Fields
• Work done 𝑊 = 𝐹𝑠
•
= 𝑞𝐸𝑑
• But
W = 𝑞∆𝑉
• Therefore 𝑞𝐸𝑑 = 𝑞∆𝑉
• ie 𝐸
=
∆𝑉
𝑑
• Thus we have a formula which gives us the electric field strength of the
uniform field between two parallel plates
• This formula gives us an alternative unit for electric field strength, the
volt per metre. 1 Vm-1 = 1 JC-1.
Example Question 3
• Calculate the strength of the uniform field is between two
parallel plates, 20cm apart, with a potential difference of 800V
between them.
Acceleration in Uniform E-Field
• If a charged particle is moving in a uniform electric field E, the
force acting on this particle is constant in magnitude and
direction, as F = qE
• Therefore the acceleration of the particle is constant in both
magnitude and direction.
• 𝑎=
𝐹
𝑚
=
𝑞𝐸
𝑚
• Note:
• If a particle is positively charged the force and acceleration are in the
same direction as the field
• If the particle is negatively charged then the force and acceleration
are in the opposite direction to the field
Motion Anti-Parallel to Field
• Motion Parallel or Antiparallel to the Field
• If the particle is initially stationary, it accelerates in the
direction of the force acting on it.
• This is identical to the motion of a body dropped in the Earth's
gravitational field
• If the particle is initially moving in the same direction as the
force acting on it, it will accelerate in the direction in which it
is initially moving.
• This is identical to the motion of a body projected vertically
downward in the Earth’s gravitational field
Motion Anti-Parallel to Field
• If the particle is initially moving in the opposite direction to
the force acting on it, it will slow down come to a halt and
then accelerate back along the path which it initially travelled.
• This is identical to the motion of a body projected vertically
upward in the Earth’s gravitational field
• In this case motion will be in two directions so take the
original direction of motion as positive
• All of theses motions are in a straight line with constant
acceleration so Newton’s equations of motion can be applied
Example Question 4
• Two large flat parallel plates have a potential difference of
200V applied to them. The plates are 10.0cm apart. An alpha
particle, initially adjacent to the positive plate, accelerates
towards the negative plate. Find:
• a) the time the proton takes to move between the plates:
• b) the final velocity of the proton (just before it hits)
• c) verify this final velocity using conservation of energy
Example Question 5
• Two large flat parallel plates, 10.0cm apart, have a potential
difference of 200V between them. A proton is fired upwards
towards the positive plate, through a hole in the negative
plate. The initial speed of the proton is 100 000 ms-1. Find:
•
•
•
•
a) the force on the proton;
b) the acceleration of the proton;
c) the time before the proton returns to the negative plate;
d) the distance the proton moves upwards in the space between
the plates.
Example Question 6
• Two parallel plates 20cm long and 5cm apart have a potential
difference of 1000V between them. In an electron gun,
electrons are accelerated through a potential difference of
4000V. On leaving the electron gun these electrons enter the
field between the parallel plates at right angles to the field, as
depicted in the diagram below. Find:
• a) the time taken for the electrons to transverse this field;
• b) the deflection of the electrons on leaving the field
Example Question 6
Application: The Cyclotron
Parts of a Cyclotron: Ion Source
• In the original cyclotron this was a heated filament
• A small quantity of hydrogen gas was introduced into the
apparatus.
• Electrons emitted by the heated filament were accelerated by
the electric field between the dees, and when they collided
with hydrogen molecules they could eject one of the hydrogen
electrons creating ions
• Modern cyclotrons have the source of ions, created by passing
a gas through an electric arc, located outside the evacuated
chamber
• These ions are then injected into the space between the dees.
• This is to preserve as good a vacuum as possible in the
apparatus
Parts of a Cyclotron: The Dees
• These are two hollow semicircular containers made of a nonmagnetic metal (eg copper)
• They are open at the diameter so that ions may pass freely
from one dee to the other.
• An alternating potential difference is applied between the
dees.
• The metal is non-magnetisable so as not to interfere with the
external magnetic field that is applied to the apparatus
Parts of a Cyclotron:
Evacuated Outer Container
• The apparatus is located in an evacuated container, also made
of non-magnetisable material
• The vacuum is maintained in the apparatus so that the ions do
not suffer energy losses due to collisions with air molecules
Operation of a Cyclotron
• An alternating potential difference ΔV volts is
applied between the two dees.
• There is an electric field between the dees but
no field inside the dees, as they are effectively
hollow charged conductors
• As the ions accelerate between the dees they
gain energy ΔE = qΔV
• Once they pass into a dee their motion is
subject only to the external magnetic field
• They are moving at right angles to this field,
and so they follow a circular path.
• When the ions exit the dees they once again
are accelerated by the electric field between
them and gain another quantum of energy ΔE
= qΔV, as the electric field has reversed its
direction (due to the alternating current)
Operation of Cyclotron
• When the ions re-enter the dee, they now
follow a path of greater radius, as they are
𝑚𝑣
travelling faster and 𝑟 =
𝑞𝐵
• Thus the ions follow a spiral path
• Each time they cross the space between
the dees they gain a quantity of energy
given by ΔE = qΔV
• Thus by being made to cross this space
many times they can be accelerated to
high energies
Example Question 7
• Cyclotrons are used in hospitals to create radioactive isotopes
of elements, which are used in research and for diagnostic
purposes. These isotopes have short half lives and need to be
created on location when about to be used.
• A cyclotron has an alternating potential difference of 2500V
applied across the dees. Protons ar emitted from the device
with an energy of 3.2MeV. Calculate:
• a) the energy gained by a proton each time it crosses the space
between the dees;
• b) the number of times N that the protons cross the space
between the dees
• a) ΔE = 2500 eV
• Or E = qΔV
•
= 1.6 x 10-19 x 2.5 x 103
•
= 4.0 x 10-19J
• b) total E = ∆𝐸 × 𝑁
• 𝑁=
𝐸𝑡
∆𝐸
• 𝑁=
3.2 ×106
2.5 ×103
• N = 1280