SACE Stage 1
Physics
Electric Fields
Introduction

Consider two charges, the force between the two
charged bodies is inversely proportional to the
square of the distance between them
1
F 2
d
Introduction

The force between two charges is directly
proportional to the strength of each charge. If
q1 and q2 represent the strength of the charges
on each of the two spheres then,
F  q1 and F  q2
Introduction

This leads to the conclusion that the force is
directly proportional to the product of the two
charges.
F  q1q2
Introduction

Combining these results, we can conclude
that the force between two charges is
proportional to the product of the two charges
and inversely proportional to the square of
the distance between the two charges.
F  q1q2
q1q2
F k 2
d
The Unit of Charge

Charges are measured in Coulombs (C).

Typically charged bodies are of the order
of mC (10-6 C).

The charge on an electron is negative and
is usually given the symbol e,
and e = 1.6 x 10-19 C.
The Unit of Charge

The charge on a proton is positive and
equal in magnitude to e.

It takes 6.25 x 1018 electrons or protons to
create a charge of one coulomb.
Coulomb’s Law

The magnitude of the electrostatic force
between the two charges is directly proportional
to the magnitudes of the charges and inversely
proportional to the square of the distance
between the charges.
q1q2
F k 2
d
Coulomb’s Law
q1q2
F k 2
r
q1 and q2 are the magnitudes of the charges on
the two bodies.
r is the distance between them.
k is the constant of proportionality,
k = 9 x 109 Nm2C-2.
Coulomb’s Law
To determine if the force is attractive or
repulsive can be determined as follows,
 Any two stationary charges experience
mutual forces along the line joining the
two centres of the charges.
 The forces are attractive if the charges
are of opposite sign and repulsive if the
two charges are of the same sign.
Coulomb’s Law
In both cases, F1 = -F2. That is the forces are
of equal magnitude but opposite in direction
(Newton’s 3rd Law).
Dependence of Force on the
Medium Between the Charges

Coulomb’s Law,
q1q2
F k 2
r
replacing k,
1
q1q2
F
2
40 r
Dependence of Force on the
Medium Between the Charges

The quantity 0 is called the permitivity of a
vacuum. Its value is 8.85 x 10-12 C2N-1m-2.

If the medium between the two charges is
replaced with oil, plastic or some other
insulating medium then the value of 0 will
need to change.
How big a charge is one Coulomb?


1 Coulomb is very large.
Consider 2 charges of 1C each, separated by a
distance of 1m in air.
q1q2
F k 2
r
1C 1C
F  9 10
1m
9
F  9 10 N
9
How big a charge is one Coulomb?

Consider 2 charges of 1C each, separated by a
distance of 1km in air.
q1q2
F k 2
r
1C 1C
F  9 10
3
10 m
6
F  9 10 N
9
The weight of a small car!
Examples
Calculate the force of repulsion between two small spheres
each carrying a charge of 4.0mC, 0.5m apart in a vacuum.
1
q1q2
F
2
40 r
9 109  4.0 10 6  4.0 10 6
F
2
(0.5)
3
9 16 10
F
0.25
F  5.96 101 N
Examples
The magnitude of the force between two charges is 5.00N.
What is the magnitude of the force if,
(a) One charge is doubled in magnitude
(b) The magnitude of one charge is multiplied by 3 and
the other is multiplied by ¼.
(c) The distance between the charges doubles.
Examples
(a) One charge is doubled in magnitude
q1 is multiplied by 2  F is multiplied by 2,
F = 10.0N
(b) The magnitude of one charge is multiplied by 3 and
the other is multiplied by ¼.
q1 x 3 and q2 x ¼  F is multiplied by 3 and by ¼, ie
¾.
F = ¾ x 5N
ie F = 3.75N
Examples
(a) The distance between the charges doubles.
If r is multiplied by 2 then F is multiplied by (1/2)2, ie
by a ¼.
F = ¼ x 5N
ie F = 1.25N
Charges in a Straight Line (1D)
If you have 3 charged spheres A, B and C and you
wish to find the force on Charge C due to the
other 2 charges you would,

Use Coulomb’s Law to find the force on C due to A.

Then repeat to find the force on C due to B.

The add the two forces vectorially to find the resultant
force on C (FC = FA + FB) .
Charges in a Straight Line (1D)
If you have 3 charged spheres A, B and C and you
wish to find the force on Charge C due to the
other 2 charges you would,

Use Coulomb’s Law to find the force on C due to A.

Then repeat to find the force on C due to B.

Then add the two forces vectorially to find the resultant
force on C (FC = FA + FB) .
Charges in a Straight Line (1D)
Two Charges, q1 and q2, of magnitude 20mC and
10mC respectively, are placed 50cm apart in a
vacuum. A third charge, q3 = +30mC, is placed
midway between q1 and q2. Find the force on q3 if,
1. q1 and q2 are positive,
2. q1 is positive and q2 is negative.
Charges in a Straight Line (1D)
Firstly determine the magnitude of the force on q3
due to q1 and q2.
q1 = 20mC = 2.0 x 10-5C, q2 = 10mC = 1.0 x 10-5C
q3 = 30mC = 3.0 x 10-5C, r = 25cm = 0.25m
1
q1q2
F
2
40 r
Charges in a Straight Line (1D)
Magnitude of force between q1 and q3,
9 109  2.0 10 5  3.0 10 5
F13 
(0.25) 2
 F13  8.64 10 N
 F13  86.4 N
Charges in a Straight Line (1D)
Magnitude of force between q2 and q3,
9 109 1.0 10 5  3.0 10 5
F23 
(0.25) 2
 F23  4.32 10 N
 F23  43.2 N
Charges in a Straight Line (1D)
(1)q1 and q2 are both positive
F13 is to the right and F23 is to the left.
 net force F  F1  F3
 43.2 N to the right
Charges in a Straight Line (1D)
(1)q1 is positive and q2 is negative
F13 and F23 are in the same direction
 net force F  F1  F2
 129.6 N to the right
Charges in a Plane (2D)
Two charges q1 and q2 of magnitude +2mC and -1mC respectively are placed
50cm apart in a vacuum. A third charge, q3 = +3mC, is placed 40cm from q1 so
as to form a right angled triangle as in the direction of the diagram. Find the
force on q3.
Charges in a Plane (2D)
1
q1q2
F1 
2
40 r
 force between q1 and q3 is given by
9 109  2.0 10 6  3.0 10 6
F1 
(0.4) 2
F1  337.5 10 3 N
F1  0.3375 N to the right
Charges in a Plane (2D)
Due to Pythagoras’
Theorem, the distance
from q2 to q3 is 30cm
(a 3-4-5 triangle).
Charges in a Plane (2D)
1
q1q2
F2 
40 r 2
 force between q2 and q3 is given by
9 109 1.0  10 6  3.0 10 6
F2 
(0.3) 2
3
F2  300.0  10 N
F2  0.300 N upwards
Charges in a Plane (2D)
By Pythagoras’ Theorem,
F 2  F12  F22
 0.33752  0.300 2
 0.2039
 F  0.2039
 0.45 N
0.3
tan  
0.3375
 0.8889
thus   41.60
Charges in a Plane (2D)
Thus the resultant force on q3 is 0.45N at an
angle of 41.6o above the horizontal.
The Electric Field

An electric field exists in a region of space
if a charged body, placed in that region,
experiences a force because of its charge.
The Electric Field

The direction of the electric field is
determined by the direction of the field
placed on a test charge in that field
Pictorial representation of the
Electric Field
Direction of Lines of Force
Electric fields are represented by using lines of
force.
Lines of force show the
direction and magnitude
of the field strength at
any point in the field.
Pictorial representation of the
Electric Field
Density of the Lines of Force
The number of Fields Lines per unit of cross sectional area
is proportional to the field strength.
If the lines of force are close together, the electric field is
stronger.
Pictorial representation of the
Electric Field
Drawing Lines of Force
1. Lines of force are always drawn away from positive
charges and towards negative charges.
2. Lines of Force always leave or come into contact at right
angles to the surface.
3. Lines of Force never cross each other.
Pictorial representation of the
Electric Field
Examples of Electric Fields
An isolated Point Charge
A positive charge
A negative charge
Pictorial representation of the
Electric Field
Examples of Electric Fields
An Electric Dipole – Two Point Charges
A positive and
negative charge
Two positive point
charges.
Pictorial representation of the
Electric Field
Examples of Electric Fields
A Charged Hollow Sphere
A positively
charged sphere
A negatively
charged sphere
Pictorial representation of the
Electric Field
Examples of Electric Fields
Two Oppositely Charged Parallel Plates
Notice that the field is curved at the ends.
Pictorial representation of the
Electric Field
Examples of Electric Fields
A Non-Uniform Conductor
The smaller the radius of curvature of a
surface, the greater the concentration of
charge in that region.
Pictorial representation of the
Electric Field
Corona Discharge
If the intensity of the electric field near a sharp projection of a conductor can
be large enough, charge can leak away from this point.
The gradual discharge is called corona discharge.
Electric Field Strength (E)
Definition
The electric field strength vector E at any point in
an electric field is defined as the force per unit
positive (test) charge placed at that point in the
field.
F
E
q
Units are NC-1
Electric Field Strength (E)
To find the electric field strength at
a point A, in the field of an isolated
point charge, +Q and at a distance
r from it, as in the diagram at the
right, place a small positive (test)
charge +q, at the point A.
Electric Field Strength (E)
The force between the 2 charges,
1
Qq
F
40 r 2
Electric field strength at A is given
by,
F
E
q
1
Q
E
40 r 2
Electric Field Strength (E)
Example,
1.
Calculate the electric field strength E at a point X,
3cm from a charge of 6mC in a vacuum.
Electric Field Strength (E)
Example,
1.
Calculate the electric field strength E at a point X,
3cm from a charge of 6mC in a vacuum.
1
Q
E
40 r 2
E 
9 109  6 10 6
3 10 
2 2
 E  6 107 NC 1 away from the charge
Electric Field Strength (E)
Example,
2.
If a charge of +10-12C is placed at X, what force
would it experience?
Electric Field Strength (E)
Example,
2.
If a charge of +10-12C is placed at X, what force
would it experience?
F
E
q
 F  Eq
 F  6 107 10 12
 F  6 10 5 N away from the 6mC charge
Electric Field Strength (E)
Example,
3.
What other information would you need if you were
to calculate the acceleration of the masses due to the
Coulombic force?
Electric Field Strength (E)
Example,
3. What other information would you need if you were to
calculate the acceleration of the masses due to the
Coulombic force?
Would need to know the masses of the bodies on which
the charges reside as,
F
a
m
Electric Field Strength (E)
Example,
4. (a) What is the electric field strength 6cm from the 6mC
charge?
Electric Field Strength (E)
Example,
4. (a) What is the electric field strength 6cm from the 6mC
charge?
As E 1/d , if d is doubled to 6cm, then E is reduced to ¼ of
its original value.
2
E = 1.5 x 107 NC-1, in a direction away from the 6mC charge.
Electric Field Strength (E)
Example,
4. (b) What is the electric field strength 1cm from the 6mC
charge?
Electric Field Strength (E)
Example,
4. (b) What is the electric field strength 1cm from the 6mC
charge?
If d is reduce to1cm (1/3 its original value) then E is
multiplied by a factor of 9.
E = 5.4x108 NC-1, in a direction away from the 6mC charge.
Superposition of Electric Fields
We can calculate the electric field strength due a point
charge using,
1
q
E
2
40 r
If we want to find the electric field strength at a point due
to one or more charges, we must calculate the individual
field strengths due to each charge and then add them
vectorially.
ET  E1  E2
Superposition of Electric Fields
Two Charges of +2.0mC and -3mC are separated by a
distance of 20cm in a vacuum.
(1)
Calculate the resultant electric field strength at X.
(2)
What would be the electric field strength at X if the 3mC charge were changed to +3mC?
Superposition of Electric Fields
(1) Calculate
the resultant electric field strength at X.
q1 = +2mC = 2 x 10-6 C
q2 = -3mC = -3 x 10-6 C
1
q
E
2
40 r
 E1  9 10 
9
2 10
6
10 
1 2
1
 E1  1.8 10 NC to the right .
6
Superposition of Electric Fields
q1 = +2mC = 2 x 10-6 C
q2 = -3mC = -3 x 10-6 C
The charge due to q2 is 1.5 times bigger than q1 due to
proportionality.
Eq
 E2  2.7  10 6 NC 1 to the right .
 ET  E1  E2
 ET  1.8  10 6  2.7  10 6
 ET  4.5  10 NC
6
1
to the right .
Superposition of Electric Fields
(b)
What would be the electric field strength at X if the 3mC charge were changed to +3mC?
If q2 is made positive then E2 is directed to the left.
ET  E1  E2
 ET  1.8  106 NC 1  (2.7  106 NC 1 )
 ET  0.9  10 NC
6
 ET  9.0  10 NC
5
1
1
to the left.
Superposition of Electric Fields
Example – Two dimensional superposition.
Two charges q1 = +2mC and q2 = -3mC are separated by a
distance of 50cm in a vacuum.
Calculate the electric field strength at the point Y, which is
40cm from q1 and 30cm from q2.
Superposition of Electric Fields
q1 = +2mC = 2 x 10-6C,
E
1
q
40 r 2
Electric Field strength due to q1,
E1  9 109
6
2 10
(0.40) 2
 E1  1.125 105 NC 1 to the right .
Electric Field strength due to q2,
2 10 6
E1  9 10
(0.30) 2
9
 E1  2.00 105 NC 1 up.
q2 = -3mC = -3 x 10-6C
By Pythagoras’ Thm,
ET2  E12  E22

E  1.125 10
2
T
  2.00 10 
5 2
ET  2.295 105 NC 1
2 105
tan  
1.125 105
  60.6o
5 2
Electric Potential

Electric Potential Energy
Any charge placed in an electric field has energy due to its
position in the field.
We call this energy – electric potential.
Electric Potential

Electric Potential Difference
Definition – The electric potential difference between two
points in an electric field is the work done per unit charge
in moving a positive (test) charge between the two points,
provided all other charges involved remain undisturbed.
workdone
potential difference 
charge
workdone
V 
q
W
V 
q
Electric Potential
The unit of electric potential difference
Work is measured in Joules and charge is measured in
Coulombs, therefore the unit is defined as,
1V  1JC
1
Electric Potential
Example
How much work is done when moving an electron across
a potential difference of 100 volts?
W  q V
W  1.6  10
19
W  1.6  10
17
 100
J
Electric Potential

Energy Changes
Whenever a charge q moves between 2 points in an
electric field, with potential difference V, its potential
energy changes according to the relation,
PE  qV
Electric Potential
The law of conservation of energy enables us to use the
one expression (qV) to determine,
The work done when a charge q moves through a
potential difference
•The change (increase or decrease) in electric potential
energy PE.
•The change (decrease or increase) in kinetic energy K.
W  PE  KE  qV
The Electron Volt as a Unit of
Energy

Definition
One electron volt (1eV) is the work done when an electron
moves through a potential difference of one volt.
Or
One electron volt (1eV) is the energy gained or lost by an
electron in moving through a potential difference of one volt.
The Electron Volt as a Unit of
Energy
Consider the following,
• If an electron - charge = -e, moves through a potential
difference of 300V, the change in potential energy is
300eV.
• If a proton - charge = +e, moves through a potential
difference of 2500V, the work done is 2500eV
The Electron Volt as a Unit of
Energy
•If an -particle - charge = +2e, moves through a potential
difference of 400V, the change in potential energy is
800eV.
• If a nitrogen ion - charge = -3e, moves through a potential
difference of 50V, the change in potential energy is 150eV.
The Electron Volt as a Unit of
Energy

Conversion between electron volts and joules.
An electron moves through a potential difference of 1 Volt.
The work done is 1 electron Volt.
Workdone  qV
 1.60 10
19
 1.60 10
19
1
J
thus 1eV  1.60 10 19 J
The Electron Volt as a Unit of
Energy
Example
In an X-ray tube electrons are accelerated across a
potential difference of 30,000V.
(1)What energy do they gain?
(2)What is their final kinetic energy?
The Electron Volt as a Unit of
Energy
(1)What energy do they gain?
loss of PE  PE  Workdone  qV
 1.6 10 19  3 10 4
 4.8 10 15 J
4.8 10 15

eV
19
1.6 10
 30,000eV
The Electron Volt as a Unit of
Energy
(2)What is their final kinetic energy?
This potential energy has been converted to kinetic
energy. Thus the final kinetic energy of the electron
is,
K  30,000eV  30keV