Physics Beyond 2000
Chapter 13 Electrostatics
http://www.sciencejoywagon.com/physicszone/lesson/07elecst/default.htm
Electrostatics
• The charges are at rest. There is not any
electrical current.
• The charges have forces on each other.
A neutral atom
• Each electron carries one unit of negative
charge.
• One unit of negative charge = -1.6 × 10-19C
A neutral atom
• The nucleus contains protons.
• Each proton carries one unit of positive
charge.
• One unit of positive charge = 1.6 × 10-19C
A neutral atom
• For a neutral atom, the number of electrons
= the number of protons.
• Total charge of a neutral atom = 0
A charged atom
• If a neutral atom gains electrons, it becomes
negatively charged.
A charged atom
• If a neutral atom loses electrons, it becomes
positively charged.
A charged atom
• If a neutral atom loses electrons, it becomes
positively charged.
Charges on insulators
• Excessive charges on insulators will stay on
the insulator.
• Example: a plastic ruler with positive
charges on it.
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+
+
+ +
+
+
+
Charges on insulated conductor
• The excess charges on the insulated
conductor remains constant unless the
conductor touches another conducting
medium.
+ +
+
+
+
+
+
+
conductor
+
insulated handle
Charges on insulated conductor
• When a conducting medium touches the
conductor, the excess charges flow away.
+ +
conducting rod
+
+
+
+
+
+
conductor
+
insulated handle
Charges on insulated conductor
• If all the charges flow away, the conductor
becomes neutral.
• This is called earthing.
conductor
conducting rod
insulated handle
Charges on insulated conductor
• In some cases, only part of the charges flow
away. Some charges still remain.
• This is called sharing.
+
+
+
+
+conducting
rod
+
conductor
+
+
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insulated handle
Charges on isolated conductor
• The conductor is completely isolated from
any other object. Its excess charges remain
constant.
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+
+
+
+
+
+
+
+
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+
Two kinds of charges
•
•
•
•
•
Positive charges
Negative charges
Unit: coulomb ,C.
One proton contains 1.6 × 10-19 C.
One electron contains - 1.6 × 10-19 C.
Two kinds of charges
• Like charges repel.
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+
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+
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Two kinds of charges
• Like charges repel.
-
-
-
-
-
-
-
-
Two kinds of charges
• Unlike charges attract.
-
-
-
- +
+
+
+
http://www.colorado.edu/physics/2000/waves_particles/wavpart2.html
Charging objects
• Use a power supply
• By rubbing
• By induction
Charging by power supply
•
•
•
•
Use a light conducting sphere and a EHT.
Ground the negative terminal of the EHT.
Let the sphere touch the positive terminal of the EHT.
What kind of charges is on the sphere?
Positive
charge
earthed
Charging by power supply
•
•
•
•
Use a light conducting sphere and a EHT.
Ground the positive terminal of the EHT.
Let the sphere touch the negative terminal of the EHT.
What kind of charges is on the sphere?
Negative charges
earthed
Charging by power supply
• Explain why the sphere is charged.
Hint: sharing the charges.
Charging by rubbing
• Use a piece of dry cloth to rub a polythene
rod.
• What kind of charges is on the polythene
rod?
Negative charges
Charging by rubbing
• Explain why the polythene rod is negatively
charged
Shuttling ball experiment
positive
negative
Shuttling ball experiment
3. An isolated light
conducting sphere.
1. Metal plate B connected
to the positive terminal
of the EHT.
2. Metal plate A connected to
the negative terminal
of the EHT.
Shuttling ball experiment
3. An isolated light
conducting sphere.
1. Connect to
the positive terminal
of the EHT.
2. Connect to
the negative terminal
of the EHT.
4. Allow the ball
to touch one plate
Shuttling ball experiment
3. An isolated light
conducting sphere.
1. Connect to
the positive terminal
of the EHT.
2. Connect to
the negative terminal
of the EHT.
5. The ball is
shuttling.
Shuttling ball experiment
Explain the experiment.
3. An isolated light
conducting sphere.
1. Metal plate connected
to the positive terminal
of the EHT.
2. Metal plate connected to
the negative terminal
of the EHT.
Charge distribution in a
conductor
• Inside the conductor: the excess charges do
not stay inside the conductor because of the
repulsion.
• On the surface of a conductor: the excess
charges reside on the surface of the
conductor.
+ +
+
+ + ++ + + + + +
+ + ++ + + + + +
+
+
+
+
+
+
+
+
+ + +
Charge distribution on the
surface of a conductor
• If the surface of the conductor is flat, the
charge distribution is uniform except at the
edge.
+ + + + + + + + + + + ++ +
+ + + + + + + + + + + ++ +
+ + + + + + + + + + + ++ +
Charge distribution on the
surface of a conductor
• If the surface of the conductor is spherical,
the charge distribution is uniform.
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+ +
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+
+
+
+
+
+
+
+ + +
Charge distribution on the
surface of a conductor
• If the surface of the conductor is of pearshaped, the charge distribution is dense on
curved surface than on flat surface.
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+
++
+
+
+
++
Charge distribution on the
surface of a conductor
• On the surface of a conductor, the surface charge
density describes the distribution of charges.
• Surface charge density  is the charge per unit
surface.
Q

A
where Q is the amount of charge on an area A.
+ +
+
+
+
+
+ + ++ + + + + +
+
+
+ + ++ + + + + +
+
+
+
+ + +
Example 2
• The surface area of a sphere is 4r2.
• The SI unit of charge Q is coulomb (C)
• The SI unit of surface charge density  is
coulomb per m2 (C m-2) .
Electric fields
• There are forces (attraction or repulsion)
between charges.
• How to describe these forces?
(How large is the force and what is its
direction if I put a charge, say 1 C, at a
point?)
• Electric field is the basic concept.
http://www.colorado.edu/physics/2000/applets/nforcefield.html
http://www.colorado.edu/physics/2000/waves_particles/wavpart3.html
Electric fields
• An electric field is a region in which an
electric charge experiences a force.
Place a charge at this point. There is
not any force on it. So the electric
field is zero at this point.
+
Electric fields
http://www.sciencejoywagon.com/physicszone/lesson/07elecst/fieldint/efield.htm
• An electric field is a region in which an
electric charge experiences a force.
Place a charge at this point. There is
force on it. So there is an electric
field here.
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+
+
+
Electric fields
• There are two ways to describe the electric
fields.
(i) Lines of forces (graphical) ;
(ii) Electric field strength (numerical).
Lines of Force
• Draw directed lines.
• Some electric field patterns:
http://www.sciencejoywagon.com/physicszone/lesson/07elecst/static/fieldmap.htm
Lines of Force
• A field line always directs away from a positive
charge and ends at a negative charge.
• The tangent to the line at any point gives the
direction of force acting on any positive test
charge.
• The number of field lines drawn per unit crosssectional area is proportional to the strength of the
electric field.
Lines of Force
• A field line always directs away from a
positive charge and ends at a negative
charge.
+
-
Lines of Force
• The tangent to the line at any point gives the
direction of force acting on any positive test
charge.
+
+
+ +
-
Lines of Force
• The force is in opposite direction if the test
charge is negative.
-
+
-
-
-
Lines of Force
• The tangent to the line at any point gives the
direction of force acting on any positive test
charge.
+
+
+
+
-
Lines of Force
• The number of field lines drawn per unit
cross-sectional area is proportional to the
strength of the electric field.
1. Weak electric field
2. Strong electric field
Electric field strength E
• The electric field strength E at a point is the
force per unit test charge placed at that point.
F
E
q
where F is the force on test charge q.
Unit of E: N C-1 or V m-1
• Example
Find the electric field strength
at the point of the charge.
q=3C
F = 18 N
Electric field strength E
• Note that the test charge q must be small.
• For a big test charge, the original electric
field may be changed.
+
+
1. The electric field of a
single charge
+
2. The electric field changes due
to the presence of another big charge.
Ionization of air by
strong electric field
1. An air molecule consist of equal number
of positive and negative charges.

forces
+ -
E
E
+
-
2. In a strong electric field, positive charges are
pulled to one side and negative charges to the
opposite side.
3. Charges are completely separated. There
forms an ion-pair.
Ionization of air by
strong electric field
• Sharp point carrying charges may create a
very strong electric field.
+
+
+
+
+ + + +++ +
+
+
+ +
• The electric field strength that causes
ionization is called the breakdown field
strength Eb. For air, Eb  106 NC-1.
Example 3
•
F
E   F = q.E
q
• Electric
force is large in comparison with
gravity.
•Effect of gravity is usually ignored when
electric force is present.
Uniform Electric field
• In a uniform electric field, the electric field
strength E is the same at every point.
• Force on a charge in a uniform electric field
is constant at every point.
F = E.q = constant
• The field lines (or lines of force) are evenly
spaced and parallel for a uniform electric field.
Uniform Electric field
• Making a uniform electric field:
1. Two parallel metal plates
with one connected to high
potential.
High potential
2. Another metal plate is
connected to low potential.
Low potential
3. Between the plates, there is a uniform
electric field
Uniform Electric field
• Making a uniform electric field:
4. The lines of force are from the plate of high potential
to the plate of low potential.
High potential
Low potential
Uniform Electric field
• Checking a uniform electric field:
5. Use a charged metal foil to test the electric field.
6. Move the foil from left to right. The angle of deflection of the
foil should not change, indicating a constant electric force.
Insulating
handle
High potential
Charged metal foil
Low potential
Uniform Electric field
• Forces on the charged foil are in equilibrium.
1.Angle of
deflection 
2. tension along the foil
+
+
3. electric force on
the charges
+
+
+
charged
foil
vertical
line
foil
4. weight of the foil
Uniform Electric field
• Forces on the charged foil are in equilibrium.
• Find FE in terms of W and .

T
FE
foil
W
FE = W.tan
Example 4
• 50 kV m-1 = 50,000 N C-1
• 2 C = 2  10-6 C
• The gravity is ignored in this example
because it is too small.
Motion of a charge
in a uniform E-field
u

q
E
y
x
1. A positive charge q is projected
with initial velocity u, making an
angle  with a uniform electric field.
2. The electric field strength is E.
3. The path of the charge will be a
parabola.
4. Prove it.
5. Hint: Separate the motion into
two, one along horizontal direction
and the other vertical.
Work done by the field
• Suppose that the charge q moves freely in a
uniform electric field E for a displacement d.
• What is its gain in kinetic energy?
• What is the work done by the field force?
W = Eqd
E
q
d
Coulomb’s Law
• Numerical method to find the force between
two charges – Coulomb’s law.
F
Q2
Q1
F
r
• For two point charges Q1 and Q2 at distance
r apart, the electric force between them is
1 Q1.Q2
F
2
4 r
Coulomb’s Law
• For two point charges Q1 and Q2 at distance
r apart, the electric force between them is
F
Q2
Q1
F
r
1 Q1.Q2
F
4 r 2
where  is the absolute permittivity of the intervening medium.
Coulomb’s Law
1 Q1.Q2
F
4 r 2
1. The absolute permittivity of free air is
o = 8.9  10-12 C2 N-1 m-2
2. For other medium, its absolute permittivity  is
expressed in terms of the relative permittivity r
and the absolute permittivity of free air o by
 = r . o .Refer to table 13-1 on page 282.
3. For free space,
1
4o
 9.0 109 Nm 2C 1
Investigate Coulomb’s law
Objective: To show that
1
F 2
r
nylon thread
L 
L
X
1. A suspended charged
sphere X of mass m.
2. The length of the nylon
thread is L.
X
d
r
Y
2. Put another charged sphere Y near
to X.
3. There is an angle of deflection .
4. Measure the separation r and d
the displacement of X .
Investigate Coulomb’s law
1. Find the electric force in terms of
m, d and L.
2. Use small angle of deflection
tan   sin 
3. This shows that F  d.
L 
X
d
r
Y
Investigate Coulomb’s law
1. Repeat the experiment by changing
the separation r. The charges should not
change.
1
2. Plot a graph of d against 2
r
L 
X
d
d
r
Y
0
1
r2
1
3. F  2
r
http://www.engr.uky.edu/~gedney/courses/ee468/expmnt/coulomb.html
Electric forces
• The electric forces act along the line joining
the two point charges.
Q
F
2
Q1
F
r
• The electric forces may be added as vectors.
F1
R
F2
Analogy to gravitational force
Coulomb’s law
Law of gravity
Inverse square law F  1
r2
Attraction or
repulsion
Negative force for attraction and
positive force for repulsion
Attraction only
Analogy to gravitational force
• http://www.pa.msu.edu/courses/1997spring/
PHY232/lectures/coulombslaw/grav%20_a
nalogy.htm
mass 1
F
r
positive charge 1
F
r
mass 2
F
negative charge 2
F
Gravitational force
is always attraction.
Electric force may be
repulsion or attraction.
Example 5
• An electron contains one unit of negative
charge Q1 = e = -1.6  10-19 C.
• An -particle has two units of positive
charge Q2 = + 2  1.6  10-19 C
• The electric force is an attractive force. So
its value is negative.
Shell Theorem
• A uniform spherical shell of charge behave,
for external points, as if all the charges were
concentrated at its centre.
+
+
+
R
+
Q1
+
+
+
1. A metal sphere with
charge Q1 uniformly
distributed.
2. Separation = r
>R
3. A point charge Q2
+
4. The force on Q2 is
1
Q1Q2
F
. 2
4o r
Shell Theorem
• A uniform spherical shell of charge exerts
no force on a charged particle placed inside
a shell.
2. Separation = r
+
Q1
+
R
+
+
r
+
<R
+
+
+
1. A metal sphere of radius Rwith
charge Q1 uniformly distributed on
its surface.
3. A point charge Q2
4. The force on Q2 is zero.
Shell Theorem
• The force between two spherical distribution
charge is equal to that of two point charges.
+
+
F
+ +
+
1. Spheres
+
Q1
r
+
+
+
F
+
Q1
+
+
+
+ + +
+
F
Q2
+
+
+
+ +
+
+
+
+ + +
2. Point charges
r
3. What is the magnitude of F?
Q2
F
Calculation of
Electric field strength (E)
• E due to a point charge +Q
The electric field strength at a
distance r from a point charge +Q is
+Q
Q
Er 
. 2
4 r
1
r
F
E
q E=?
q
The field is spherically symmetrical.
Example 6
• Is an alpha-particle a point charge?
• What is the quantity of charge of an alphaparticle?
• What is the distance 1 m?
• The unit of E is either NC-1 or Vm-1.
Calculation of
Electric field strength (E)
• E due to a charged spherical metal shell.
1. The charges reside on the surface of the
spherical shell.
2. The E inside the shell is 0.
3. The E outside the shell is similar to that of a
point charge at the centre of the shell.
+ + +
+
+
+
+
+
+
+
+ + +
Calculation of
Electric field strength (E)
• E due to a charged spherical metal shell.
1. E = 0 inside the
conducting shell.
R
3. What is E just
on the surface of
the shell?
Q
r
2. Outside the shell
1
Q
Er 
. 2
4 r
Calculation of
Electric field strength (E)
• E due to a charged spherical metal shell.
4. Use the surface
charge density 
to simplify the
answer.
R
3. What is E just
on the surface of
the shell?
Q

ER 

Calculation of
Electric field strength (E)
• E due to a charged spherical metal shell.
1. E = 0 inside the
conducting shell.


E
0
R

2. On the surface E R 

r
R
3. Outside the shell
R
1
Q
Q
Er 
. 2
4 r
Calculation of
Electric field strength (E)
• E on the surface of any charged conductor

E

Note the  may be different on different positions of the
non-spherical conductor.
 is small on a flat surface and high on curved surface.
Calculation of
Electric field strength (E)
• E on the surfaces and between two infinite large
and parallel charged sheets
1. Positively
charged plate

E

2. Positively
charged plate
3. Constant electric
field strength between
Example 7
• The medium between the parallel plates is
air.

• Use E 
where o = 8.9  10-12 C2 N-1 m-2
0
Calculation of
Electric field strength (E)
• E due to a uniform spherical charge.
+
++
+ +
++
+
+ +
+ + +
+ + +
+ + +
+ + +
+ + +
+ +
+
++
+ +
+
+
1. charges are uniformly
distributed in the sphere
+
+
+
+
2. Charges are on
the surface only.
+
+
+
+ A metal shell
or metal sphere
Calculation of
Electric field strength (E)
• E due to a uniform spherical charge.
+
++
+ +
++
+
Radius = R
+ +
+ + +
+ + +
+ + +
+ + +
+ + +
+ +
+
++
+ +
+
+
1. Total number of charges = Qo
r
2. Outside the sphere at a
distance r from the centre
Qo
Er 
. 2
4 r
1
3. Inside the sphere at a distance r
from the centre
Qo .r
Er 
. 3
4 R
1
Calculation of
Electric field strength (E)
• E due to a uniform spherical charge.
Qo
1. Er 
. 2
4 r
1
R
+
++
+ +
++
+
0
+ +
+ + +
+ + +
+ + +
+ + +
+ + +
+ +
r
R
+
++
+ +
+
+
Qo .r
2. Er 
. 3
4 R
1
Gauss’ Law
• Gauss’ law is a convenient way to calculate
the electric field strength E, especially when
Coulomb’s law cannot be applied.
Gauss’ law
• Definition of Electric flux  :
The electric flux  across a surface area is
the product of the surface area A and the
normal component of the electric field
E
strength E

 = E . A
A
Example 8
• Note that we have to calculate the E in the
second part.
E
E

A
Find the E and thus  from the right diagram.
A
Gauss’ law
• The total electric flux normal to a closed
Q
imaginary surface is equal to


where Q is the charge enclosed by an area
and  is the permittivity
1. Suppose that
there is a charge
Q.
Q
2. Draw an imaginary
surface to enclose the
charge
3. The electric flux
through this surface
is always   Q

Gauss’ law
• Use Gauss’ law to find the electric field strength
at a distance r from a point charge Q.
Q
r
1. Draw a spherical surface
of radius r around the charge Q.
2. What is the total electric flux?
3. Find E from  = E .A.
Note that E is  the surface.
1
Q
E  Er 
. 2
4 r
Gauss’ law
• Use Gauss’ law to find the electric field strength at a
distance r from a line of charge with charge per unit
length .
• Note that it is difficult to use Coulomb’s law here.
+
Er = ?
x
r
+
+
+
+
+
+
+
Gauss’ law
• Use Gauss’ law to find the electric field strength at a
distance r from a line of charge with charge per unit
length .
+
Er = ?
x
r
+
+
+
+
+
+
+
h
1. Draw a cylindrical surface
of radius r and height h around the
line of charge.
2. What is the total electric flux?
3. Find E from  = E .A. Note
that E is  the surface.

E  Er 
2r
Gauss’ law
• Find the electric field strength E due to a
hollow charged conductor enclosing a point
charge.
a
+Q1
b
+Q2
Gauss’ law
• In the cavity of the hollow sphere with
0 < r  b.
a
1. Draw a spherical surface
of radius r around the charge Q1.
2. What is the total electric flux?
3. Find E from  = E .A.
Note that E is  the surface.
+Q1
b
+Q2
1
Q1
E1 
. 2
4 r
Gauss’ law
• Inside the conducting sphere with b  r  a.
a
+Q1
-Q1
+Q2+Q1
1. Draw a spherical surface of radius
r inside the conducting sphere.
2. What is the charge enclosed by
this spherical surface?
3. What is the total electric flux?
4. Find E from  = E .A. Note that
E is  the surface.
b
-Q1
+Q1+Q2
E2 = 0
Gauss’ law
• Outside the hollow sphere with r  a.
a
+Q1
b
+Q2
1. Draw a spherical surface
of radius r around the charges.
2. What is the charge enclosed?
3. What is the total electric flux?
4. Find E from  = E .A.
Note that E is  the surface.
Q1  Q2
E3 
.
4
r2
1
E-r Graph
1
Q1
E1 
. 2
4 r
a
+Q1
b
+Q2
E2 = 0
Q1  Q2
E3 
.
2
4
r
E
a
b
0
1
b
a
r
Electric potential
• Every test charge in an electric field
possesses electric potential energy.
• The best way to describe the distribution of
energy is using electric potential V.
• If the electric potential at a point in an
electric field is V and we place a test charge
q at the point, then the electric energy is
U = q.V
Electric potential
• Work done on a test charge +q in a uniform
electric field E.
• Move a test charge +q from A to B against
the electric force. We need to apply an
external force F and the work done = F.d
2. Electric field strength E
1.External force = F
+q
A
d
B
3. Displacement
AB = d.
Electric potential
• Find the least work done W.
W = qEd
• The charge +q has gained an electric potential
energy U = qEd.
2. Electric field strength E
1.External force = F
+q
A
d
B
3. Displacement
AB = d.
Electric potential
• The charge +q has gained an electric potential
energy U = qEd.
• What would happen to the charge if it is released
at B?
1. Electric field strength E
2. Charge +q
B
Electric potential
• Electric potential V at a point is the work done per
coulomb required to bring a positive charge from
infinity to the point.
• The electric potential at infinity is zero.
2. A varying force F to bring the
charge from infinity to this point X.
1. A +q charge at ∞
F
X
3. Work done on the charge = W
4. The electric potential energy of the charge U = W.
5. Electric potential at X is V  U
q
∞
Electric potential V
• In practice, the electric potential at a point is
often referred to earth rather than infinity.
• In practice, the electric potential on the
ground is zero.
Electric potential V
• The unit of V is J C-1 (joule per coulomb) or
V (volt).
• The electric potential energy U of a charge
q at a point of electric potential V is
U = q.V
Electric potential difference ΔV
• ΔVBA = VB - VA
Electric potential energy difference ΔU
• ΔUBA = UB - UA
Electric potential difference ΔV
• ΔVBA = VB – VA
• The potential between two points is the
work done per coulomb on a positive charge
in moving between them.
ΔV
U

q
Example 8
• Note that the direction of motion.
ΔVAB = - ΔVBA
• If ΔU > 0, there is a gain in electric
potential energy.
If ΔU < 0, there is a loss in electric potential
energy.
• What would happen if the proton is allowed
to move freely?
Potential difference of
parallel plates
• The potential difference V, the electric field E and
the separation of the parallel plates are closed
related.
1. Set the
potential
to be zero.
2. Potential = V
4. Separation = d
d
+
3. Electric field
strength = E
Potential difference of
parallel plates
• Find the potential difference between plates.
2. Move the charge to
the other plate. What
is the minimum force
F needed?
1. Place a charge +q
on the side with 0V.
-
d
V  E.d 

+
d
3. Find the work done
and calculate V.
Potential between
parallel plates
• Find the potential difference between plates.
2. Move the charge to
a point with distance r
from the plate. What is
the minimum force F
needed?
1. Place a charge +q
on the side with 0V.
-

V  E.r 
.r

r
+
d
3. Find the work done
and calculate V.
Potential between
parallel plates
• Find the potential at any point between parallel
plates.

Vr  E.r  .r

1.
2. V changes
linearly between
the parallel plates
d
V  E.d 

3.
V
0
d
-
r
+
d
electronvolt (eV)
• 1 eV = work done on an electron in moving
it through a potential difference of 1 V.
• It is a unit of energy.
• It is commonly used to measure the energy
of particle
Example 9
• Ignore the gravitational force. It is too small
in comparison with the electric force.
electronvolt (eV)
• Find the absolute value of 1 eV in terms of
the unit J.
1. What is the charge q of
an electron?
ΔV = 1V
4. 1 eV = 1.6 × 10-19 J
2. Move it through a
potential difference of 1 V.
3. Calculate U from U = qV.
Potential due to a point charge
1. There is a point
charge Q.
2. Take a charge q at
infinity.
q
Q
1
Q
V
.
4 r
∞
r
3. Bring the charge q to
this position at a distance
r from the charge Q.
4. Calculate the work
done W and V.
Potential due to a point charge
Q
V
0
1
Q
V
.
4 r
r
Example 10
• 1 kV = 1000 V
• The potential may either be positive or
negative depending on the point charge Q.
Potential due to a
charged metal sphere
• The
charge Q of the sphere resides on the surface.
3. Constant potential inside
the metal sphere!
1. Outside the sphere, the charged
sphere is similar to a point charge.
What is the potential at this point?
+ + +
+
R +
+ Q
r
+
+
+
+
+ + +
2. What is the
potential on the
VR
surface of the sphere?
1
Q
Vr 
.
4 r
1
Q

.
4 R
Potential due to a
charged metal sphere
1.
1
Q
VR 
.
4 R
+ + +
+
R +
+
+
Q
+
+
+
+ + +
V
2.
Q
Vr 
.
4 r
R
0
R
1
r
Example 11
• Note that the potential of a metal sphere is
constant inside in electrostatics.
• Two cases, outside or inside the sphere,
must be considered.
Electric field strength
and potential
dV
E
dr
• Field strength = rate of change of potential
• The negative sign indicates the direction
of electric field strength
• Electric field strength is also called potential
gradient.
Electric field strength
and potential
• Find the electric field strength due to a point
charge
1 Q
V
.
4 r
• Find the electric field strength due to a
charged metal sphere
Inside the sphere
1
Q
VR 
.
4 R
Outside the sphere
1
Q
Vr 
.
4 r
Equipotential surfaces
• An equipotential surface is an imaginary
surface on which the potential is the same.
• Work done is zero to move a test charge on
the equipotential surface.
• Electric field lines are always normal to the
equipotential surfaces.
Equipotential surfaces
Plotting equipotential lines
Example 12
V
E
r
Note that E = 0 if V is constant
e.g. inside a metal sphere.
Influence on the potential
of a charged object
• The potential of a charged object is affected by
objects nearby.
• We study an example of a positively charged
conducting sphere A.
+ + +
+
R +
+
+
+
+
A
+
+ + +
Influence on the potential
of a charged object
• By an earthed conductor
+ + +
+
R +
+
1. The potential
+
A
+
without the earthed ++
+
+
+
conductor
V
4. The negative charges
reduces the potential.
The potential with the
earthed conductor.
0
2. By induction, there
are negative charges
-3. The potential of
the earthed conductor
is zero
r
Influence on the potential
of a charged object
• By an uncharged conductor
+ + +
1. The potential +
R +
+
without the
+
A
+
+
uncharged
+
+ + +
conductor
V
2. By induction, there
are negative and positive charges
--
+ +
+
+
+
4. The potential of the
conductor is constant.
3. The potential with
the uncharged conductor
0
r
Influence on the potential
of a charged object
• By an uncharged conductor
+ + +
1. The potential +
R +
+
without the
+
A
+
+
uncharged
+
+ + +
conductor
V
--
+ +
+
+
+
3. The positive induced
charges raises the potential..
2. The negative charges
reduces the potential.
0
r
Influence on the potential
of a charged object
• By a positively charged
conductor
+ + +
1. The potential +
R +
+
without the
+
A
+
charged conductor +
+
+ + +
V
3. The potential with
the uncharged conductor.
The potential is raised
by the positive charges
on the conductor.
0
2. The positive charges are
repelled to the far side.
+ +
+
+
+
4. The potential of the
conductor is constant.
r
Influence on the potential
of a charged object
• By a negatively charged
conductor
+ + +
1. The potential +
R +
+
without the
+
A
+
charged conductor +
+
+ + +
V
3. The potential with
the uncharged conductor.
The potential is reduced
by the negative charges
0
on the conductor.
2. The negative charges are
attracted to the near side.
--
4. The potential of theconductor
is constant and negative.
r
Influence on the potential
of a charged object
• By a flame probe (an isolated conductor in flame)
+ + +
1. The potential +
R +
with and without + A
+
+
the flame probe. +
+
+ + +
V
0
2. The flame
probe is neutral
and does not
affect the potential
r
Influence on the potential of
charged parallel plates
d
0V
+Vo
1. A pair of parallel
plates at constant
potential difference
and separation d.
V
Vo
2. The change of potential
inside is linear.
r
0
d
Influence on the potential of
charged parallel plates
d
1. Place a metal block
of thickness b between the
plates
0V
b
2. The potential of the
metal block is a constant.
+Vo
V
Vo
Vo
2
3. If the block is exactly
at the
V middle, its potential
is 2
o
r
Flame probe
Flame probe
1. By induction, there are induced charges (including negative
and positive charges) on the isolated conductor.
2. The flame produces ions (including positive and negative
ions) to neutralize the induced charges.
+ + +
+
R +
+
+
+
+
+
+ + +
+
+
3. It is neutral
-
2. Ions produced by
the flame
+
+
1. Induced charges
Flame probe investigation
• As a flame probe does not affect the
potential, we can use it to measure the
potential at a point.
• The potential at the tip of the needle is
shown from the angle of deflection of the
gold leaf. The higher the potential, the
bigger the angle of deflection.
metal
rod
θ
gold-leaf
Flame probe investigation
Flame probe investigation
• Move the flame probe away from the
conducting sphere.
•
1
V
outside a charged metal sphere
r
connected to gold-leaf
electroscope
Metal sphere
at high potential
Flame probe investigation
• What would be the result if the probe is not
in flame ?
(Hint: The probe is an uncharged conductor
if it is not in flame.)
Flame probe investigation
• Move the flame probe between two parallel
plates.
• V r
-
+
connected to gold-leaf
electroscope
Ice-pail experiment
A + +
+
B
1. A positively charged
metal sphere A.
2. A neutral metal container
B with a narrow mouth
3. A gold-leaf electroscope
to measure the charges
Ice-pail experiment
A + +
+
4. The container B
and the electroscope
are neutral. The gold-leaf
does not deflect.
B
1. A positively charged
metal sphere A.
2. A neutral metal container
B with a narrow mouth
3. A gold-leaf electroscope
to measure the charges
Ice-pail experiment
1. Lower the positively charged
metal sphere A into the container
B without touching it.
+-
3. A gold-leaf deflects
because of the induced
charges.
+
-A+ ++
- + -+
+ B
- - +
+
+
2. Induced negative charges
inside and equal amount of
induced positive charges
outside.
Ice-pail experiment
1. Let A touch the inside of B.
+
+
+
+
B
+
A
+
2. The metal sphere loses
all its charges. The induced
charges are also neutralized.
+
+
3. The deflection of
gold-leaf does not change
because the induced positive
charges do not change.
Ice-pail experiment
1. Move A away. It is
neutral.
A
+
+
+
+
+
+
B
+
+
2. All charges of A has
been transferred to B
and the electroscope.
3. The deflection of
gold-leaf does not change
because the induced positive
charges do not change.
Ice-pail experiment
We may conclude that when we put a charge to touch the insi
of a metal container, the charge will distribute completely to
the outside surface of the metal container.
+
+
+
+
+
+
B
A
+
+