PHY 2049: Physics II
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PHY 2049: Physics II
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About bicycle riding: If some one shows
you a trick, it means that,
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It is possible to do that trick
The person who showed you, can do the trick.
It does not mean that you can do the trick.
Try it a lot of times and you can too.
PHY 2049: Physics II
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U = k q1 q2/r
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Force => work =>
change in K=>
change in Potential
energy
ΔU = Uf – Ui = -W =
- ΔK
Work done is path
independent. K+U =
constant.
PHY 2049: Physics II
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Electric Potential
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V = U/q = -W/q
Units of Joules/coulomb = volt
1 eV = e x 1V = 1.6 x 10-19 J
Also
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V = kq/r
Vf –Vi = -∫E.ds
Add as a number
PHY 2049: Physics II
PHY 2049: Physics II
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Rank them by the magnitude of electric field
Which has the field pointed to the right
What does an electron do when released midway
PHY 2049: Physics II
PHY 2049: Physics II
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Vf-Vi = -∫ k q/r2 dr
Choose Vi = V (∞)=0
V(r) = kq/r
PHY 2049: Physics II
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V = kpcosθ/r2
E = -∂V/∂s = Uniformly charged disk
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V = ??
Example : Potential due to an electric dipole
Consider the electric dipole shown in the figure
We will determine the electric potential V created at point P
by the two charges of the dipole using superposition.
Point P is at a distance r from the center O of the dipole.
p
Line OP makes an angle  with the dipole axis
1  q
q 
q r(  )  r(  )
V  V(  )  V(  ) 



4 o  r(  ) r(  )  4 o r(  ) r(  )
We assume that r d where d is the charge separation
From triangle ABC we have: r(  )  r(  )  d cos 
Also: r(  ) r(  )
where p  qd  the electric dipole moment
A
C
B
q d cos 
1 p cos 
 r V 

4 o r 2
4 o r 2
2
V
p cos 
4 o r 2
1
(24 - 6)
r
dq
V
1
dq
4 o  r
Potential due to a continuous charge distribution :
P Consider the charge distribution shown in the figure
In order to determine the electric potential V created
by the distribution at point P we use the principle of
superposition as follows:
1. We divide the distribution into elements of charge dq
For a volume charge distribution dq   dV
For a surface charge distribution dq   dA
For a linear charge distribution dq   d
.
2. We determine the potential dV created by dq at P
dV 
1
dq
4 o r
3. We sum all the contributions in the form of the integral: V 
1
dq
4 o  r
Note 1 : The integral is taken over the whole charge distribution
Note 2 : The integral involves only scalar quantities
(24 - 7)
Example : Potential created by a line of charge of
length L and uniform linear charge density λ at point P.
Consider the charge element dq   dx at point A, a
distance x from O.
From triangle OAP we have:
r  d 2  x 2 Here d is the distance OP
The potential dV created by dq at P is:
1 dq
1
 dx
dV 

4 o r
4 o d 2  x 2
O
dq
A

V
4 o

L

0
dx
d 2  x2

V
4 o

V
4 o
dx
d 2  x2

 ln x  d 2  x 2

ln L 
 


L  x   ln d 

L
ln x  d  x 

 0
2
2
2
2
(24 - 8)
(24 - 9)
Induced dipole moment
Many molecules such as H 2O have a permanent electric
dipole moment. These are known as "polar" molecules.
Others, such as O 2 , N 2 , etc the electric dipole moment
is zero. These are known as "nonpolar" molecules
One such molecule is shown in fig.a. The electric dipole
moment p is zero because the center of the positive
charge coincides with the center of the negative charge.
In fig.b we show what happens when an electric field E
is applied on a nonpolar molecule. The electric forces
on the positive and nagative charges are equal in magnitude
F(  )
F(  )
but opposite in direction
As a result the centers of the positve and negative charges move in opposite
directions and do not coincide. Thus a non-zero electric dipole moment p
appears. This is known as "induced" electric dipole moment and the molecule
is said to be "polarized". When the electric field is removed p disappears
W  qV
Equipotential surfaces
A collection of points that have the same
potential is known as an equipotential
surface. Four such surfaces are shown in
the figure. The work done by E as it moves
a charge q between two points that have a
potential difference V is given by:
W  qV
For path I : WI  0 because V  0
For path II: WII  0 because V  0
For path III: WIII  qV  q V2  V1 
For path IV: WIV  qV  q V2  V1 
Note : When a charge is moved on an equipotential surface  V  0 
The work done by the electric field is zero: W  0
(24 10)
The electric field E is perpendicular
to the equipotential surfaces
Consider the equipotential surface at
potential V . A charge q is moved
E
V
F

A
q
B
r
by an electric field E from point A
S
to point B along a path r .
Points A and B and the path lie on S
Lets assume that the electric field E forms an angle  with the path r .
The work done by the electric field is: W  F  r  F r cos   qE r cos 
We also know that W  0. Thus: qE r cos   0
q  0, E  0, r  0 Thus: cos   0    90
The correct picture is shown in the figure below
E
S
V
(24 11)
Examples of equipotential surfaces and the corresponding electric field lines
Uniform electric field
Isolated point charge
Electric dipole
Equipotential surfaces for a point charge q :
q
q
V
Assume that V is constant  r 
 constant
4 o r
4 oV
Thus the equiptential surfaces are spheres with their center at the point charge
q
and radius r 
4 oV
(24 -
(24 13)
Calculating the electric field E from the potential V
Now we will tackle the reverse problem i.e. determine E if we know V .
Consider two equipotential surfaces that corrspond to the values V and V  dV
separated by a distance ds as shown in the figure. Consider an arbitrary direction
represented by the vector ds . We will allow the electric field
to move a charge qo from the equipotenbtial surface V to the surface V  dV
A
B
V
V+dV
The work done by the electric field is given by:
W  qo dV (eqs.1)
also W  Fds cos   Eqo ds cos  (eqs.2)
If we compare these two equations we have:
dV
Eqo ds cos   qo dV  E cos   
ds
From triangle PAB we see that E cos  is the
component Es of E along the direction s.
Thus: Es  
V
s
Es  
V
s
Es  
V
s
We have proved that: Es  
V
s
(24 14)
The component of E in any direction is the negative of the rate
at which the electric potential changes with distance in this direction
A
B
V
V+dV
If we take s ro be the x- , y -, and z -axes we get:
V
Ex  
x
V
Ey  
y
V
Ez  
z
If we know the function V ( x, y, z )
we can determine the components of E
and thus the vector E itself
E
V ˆ V ˆ V ˆ
i
j
k
x
y
z
(24 15)
Potential energy U of a system of point charges
We define U as the work required to assemble the
q2
y
r12
system of charges one by one, bringing each charge
from infinity to its final position
Using the above definition we will prove that for
r23
q1
r13
O
a system of three point charges U is given by:
q3
x
q2 q3
q1q3
q1q2
U


4 o r12 4 o r23 4 o r13
Note : each pair of charges is counted only once
For a system of n point charges
U
1
4 o
n

i , j 1
i j
qi q j
rij
qi  the potential energy
U is given by:
Here rij is the separation between qi and q j
The summation condition i  j is imposed so that, as in the
case of three point charges, each pair of charges is counted only once
y
Step
1

q1
x
O
Step
2
y
q1

r12
q2
(24 16)
Step 1 : Bring in q1
W1  0
(no other charges around)
Step2 : Bring in q2
W2  q2V (2)
V (2) 
q1
4 o r12
 W2 
q1q2
4 o r12
Step3 : Bring in q3
x
O
y
Step
3
1  q1 q2 
  
4 o  r13 r23 
1  q1q3 q2 q3 
W3 



4 o  r13
r23 
V (3) 
r12
q2
q1

r23
r13
q3
O
W3  q3V (3)
x
W  W1  W2  W3
q2 q3
q1q3
q1q2
W


4 o r12 4 o r23 4 o r13
(24 17)
conductor
path
B

E 0
A
Potential of an isolated conductor
We shall prove that all the points on a conductor
(either on the surface or inside) have the same
potential
A conductor is an equipotential surface
Consider two points A and B on or inside an conductor. The potential difference
VB  VA between these two points is give by the equation:
B
VB  VA    E  d S
A
We already know that the electrostatic field E inside a conductor is zero
Thus the integral above vanishes and VB  VA for any two points
on or inside the conductor.
Isolated conductor in an external electric field
We already know that the surface of a conductor
is an equipotential surface. We also know that
the electric field lines are perpendicular to the
equipotential surfaces.
From these two statements it follows that the electric field vector E is
perpendicular to the conductor surface, as shown in the figure.
All the charges of the conductor reside on the surface and arrange
themselves in such as way so that the net electric field inside the
conductor Ein  0.
The electric field just out side the conductor is: Eout 

o
(24 18)
(24 19)
E
n̂
Eout 

nˆ
o
E
Ein  0
Electric field and potential
in and around a charged
conductor. A summary
n̂
1. All the charges reside on the conductor surface.
2. The electric field inside the conductor is zero
Ein  0
3. The electric field just outside the conductor is: Eout 

o
4. The electric field just outside the conductor is perpendicular
to the conductor surface
5. All the points on the surface and inside the conductor have the same potential
The conductor is an eequipotential surface
Electric field and electric potential
for a spherical conductor of radius R
and charge q
For r  R , V 
(24 20)
1
q
4 o R
1
q
For r  R , V 
4 o r
R
1
For r  R ,
q
E
4 o R 2
For r  R ,
E
1
q
4 o r 2
Note : Outside the spherical conductor the electric field
and the electric potential are identical to that of a point
charge equal to the net conductor charge and placed
at the center of the sphere