5/16/2011
PHY 2049: Physics II
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Quiz 2
The electric field at a distance of 1 m
from an isolated point particle with a
charge of 2 nC is:
A. 1.8N/C
B. 180N/C
C. 18N/C
D. 1800N/C
E. none of these
PHY 2049: Physics II
An isolated charged point particle produces
an electric field with magnitude E at a point
2m away from the charge. A point at which
the field magnitude is 4E is:
A. 1m away from the particle
B. 0.5m away from the particle
C. 2m away from the particle
D. 4m away from the particle
E. 8m away from the particle
Potential Energy and Potential
U = k q1 q2/r :
interaction energy of two
charges. Sign matters
quiz
Force => work => change
in K=> change in
Potential energy
∆U = Uf – Ui = -W = - ∆K
Work done is path
independent.
K+U = constant.
Last week
Coulomb’s law, Electric Field and Gauss’
theorem
Today
Electric Potential Energy and Electric
Potentials
Numerous cases
PHY 2049: Physics II
Electric Potential
V = U/q = -W/q
Units of Joules/coulomb = volt
1 eV = e x 1V = 1.6 x 10-19 J
Also
V = kq/r
Vf –Vi = -∫E.ds
In case of multiple charges, add as a
number
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5/16/2011
PHY 2049: Physics II
Vf-Vi = -∫ k q/r2 dr
Choose Vi = V (∞)=0
V(r) = kq/r
V = kpcosθ/r2
E = -∂V/∂s = Uniformly charged disk
V = ??
Example : Potential created by a line of charge of
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
length L and uniform linear charge density λ at point P.
Consider the charge element dq = λ dx at point A, a
distance x from O. From triangle OAP we have:
by the two charges of the dipole using superposition.
Point P is at a distance r from the center O of the dipole.
r
p
r = d 2 + x 2 Here d is the distance OP
The potential dV created by dq at P is:
Line OP makes an angle θ with the dipole axis
 q
q 
q r( − ) − r( + )
−

 =
 r( + ) r( − )  4πε o r( − ) r( + )
We assume that r d where d is the charge separation
From triangle ABC we have: r( − ) − r( + ) ≈ d cos θ
V = V( + ) + V( − ) =
Also: r( − ) r( + ) ≈ r 2 → V ≈
A
C
dV =
1
4πε o
O
dq
V=
A
∫
d cos θ
1 p cos θ
=
4πε o r 2
4πε o r 2
q
p cos θ
V=
4πε o r 2
(24 - 6)
.
P
PHY 2049: Physics II
L
∫
0
dx
λ dx
d 2 + x2
dx
d 2 + x2
(
= ln x + d 2 + x 2
(
(
)
)
)
λ 
ln x + d 2 + x 2 
 0
4πε o 
L
λ 
ln L + L2 + x 2 − ln d 
V=

4πε o 
1
B
λ
4πε o
d 2 + x2
V=
where p = qd = the electric dipole moment
1 dq
1
=
4πε o r
4πε o
r
dq
(24 - 8)
Potential due to a continuous charge distribution :
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
1
dq
V=
4πε o ∫ r
For a volume charge distribution dq = ρ dV
For a surface charge distribution dq = σ dA
For a linear charge distribution dq = λ d l
2. We determine the potential dV created by dq at P
dV =
1 dq
4πε o r
3. We sum all the contributions in the form of the integral: V =
1
dq
4πε o ∫ r
Note 1 : The integral is taken over the whole charge distribution
Note 2 : The integral involves only scalar quantities
(24 - 7)
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5/16/2011
W = − q∆V
Induced dipole moment
(24 - 9)
Equipotential surfaces
A collection of points that have the same
Many molecules such as H 2 O have a permanent electric
dipole moment. These are known as "polar" molecules.
Others, such as O2 , N 2 , etc the electric dipole moment
potential is known as an equipotential
surface. Four such surfaces are shown in
r
the figure. The work done by E as it moves
a charge q between two points that have a
is zero. These are known as "nonpolar" molecules
One such molecule is shown in fig.a. The electric dipole
r
moment p is zero because the center of the positive
potential difference ∆V is given by:
W = −q∆V
charge coincides with the center of the negative charge.
r
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
r
F( + )
r
F( − )
but opposite in direction
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 )
As a result the centers of the positve and negative charges move in opposite
r
directions and do not coincide. Thus a non-zero electric dipole moment p
The work done by the electric field is zero: W = 0
appears. This is known as "induced" electric dipole moment and the molecule
r
is said to be "polarized". When the electric field is removed p disappears
r
The electric field E is perpendicular
r
E
r
F
q
Isolated point charge
Uniform electric field
Consider the equipotential surface at
potential V . A charge q is moved
r
by an electric field E from point A
r
S
to point B along a path ∆r .
B
r
∆r
Examples of equipotential surfaces and the corresponding electric field lines
to the equipotential surfaces
V
θ
A
(24 10)
Electric dipole
Points A and B and the path lie on S
r
r
Lets assume that the electric field E forms an angle θ with the path ∆r .
r r
The work done by the electric field is: W = F ⋅ ∆r = F ∆r cos θ = qE ∆r cos θ
We also know that W = 0. Thus: qE ∆r cos θ = 0
q ≠ 0, E ≠ 0,
Equipotential surfaces for a point charge q :
∆r ≠ 0 Thus: cos θ = 0 → θ = 90°
V=
The correct picture is shown in the figure below
r
E
(24 11)
r
Calculating the electric field E from the potential 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
r
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
Es = −
∂V
∂s
We have proved that: Es = −
A
also W = Fds cos θ = Eqo ds cos θ (eqs.2)
If we compare these two equations we have:
dV
Eqo ds cos θ = −qo dV → E cos θ = −
ds
From triangle PAB we see that E cos θ is the
r
component Es of E along the direction s.
∂V
∂s
Es = −
∂V
∂s
∂V
∂s
(24 14)
r
The component of E in any direction is the negative of the rate
at which the electric potential changes with distance in this direction
The work done by the electric field is given by:
W = − qo dV (eqs.1)
Thus: Es = −
q
= constant
4πε oV
(24 12)
(24 13)
r
Now we will tackle the reverse problem i.e. determine E if we know V .
Assume that V is constant → r =
Thus the equiptential surfaces are spheres with their center at the point charge
q
and radius r =
4πε oV
S
V
q
4πε o r
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 )
r
we can determine the components of E
r
and thus the vector E itself
r
∂V ˆ ∂V ˆ ∂V ˆ
E=−
i−
j−
k
∂x
∂y
∂z
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5/16/2011
y
(24 15)
Potential energy U of a system of point charges
We define U as the work required to assemble the
system of charges one by one, bringing each charge
from infinity to its final position
Using the above definition we will prove that for
q2
y
r12
r23
q1
r13
O
U=
1
4πε o
qi q j
n
∑
q1
r12
y
r12
A
W=
q1q2
qq
qq
+ 2 3 + 1 3
4πε o r12 4πε o r23 4πε o r13
Isolated conductor in an external electric field
We already know that the surface of a conductor
is an equipotential surface. We also know that
A conductor is an equipotential surface
A
r
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.
n̂
1  q1q3 q2 q3 
+


r23 
4πε o  r13
W = W1 + W2 + W3
the electric field lines are perpendicular to the
equipotential surfaces.
VB − VA between these two points is give by the equation:
B
r ur
VB − VA = − ∫ E ⋅ d S
r
E
 q1 q2 
 + →
 r13 r23 
x
potential
Consider two points A and B on or inside an conductor. The potential difference
(24 19)
1
4πε o
W3 =
q3
(either on the surface or inside) have the same
B
∞
r23
r13
We shall prove that all the points on a conductor
r
E =0
q2
q1
Potential of an isolated conductor
path
W3 = q3V (3)
V (3) =
Step
3
O
conductor
q1
qq
→ W2 = 1 2
4πε o r12
4πε o r12
V (2) =
x
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
(24 17)
W2 = q2V (2)
q2
Step3 : Bring in q3
U is given by:
Here rij is the separation between qi and q j
rij
i , j =1
i< j
{qi } the potential energy
(no other charges around)
Step2 : Bring in q2
∞
O
Note : each pair of charges is counted only once
For a system of n point charges
Step
2
y
qq
qq
qq
U= 1 2 + 2 3 + 1 3
4πε o r12 4πε o r23 4πε o r13
x
W1 = 0
x
O
(24 16)
Step 1 : Bring in q1
∞
q1
a system of three point charges U is given by:
q3
Step
1
r
σ
Eout = nˆ
εo
r
E
r
Ein = 0
r
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 =
Electric field and electric potential
Electric field and potential
in and around a charged
for a spherical conductor of radius R
1. All the charges reside on the conductor surface.
Ein = 0
3. The electric field just outside the conductor is: Eout =
σ
εo
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
(24 18)
(24 20)
and charge q
conductor. A summary
n̂
2. The electric field inside the conductor is zero
σ
εo
R
For r < R ,
V=
1 q
4πε o R
For r > R ,
V=
1 q
4πε o r
For r < R ,
E=
1 q
4πε o R 2
For r > R , E =
1 q
4πε 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
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5/16/2011
PHY 2049: Physics II
1.
2.
3.
4.
5.
Neg., lower
?
Pos., higher
Neg., higher
Pos., higher
Can you tell the sign of the charge by looking at its
behavior over a surface of potentials. Is the speed at
the end bigger or smaller than in the beginning.
5