Lecture 4 Electric Potential
and/ Potential Energy Ch. 25
•Review from Lecture 3
•Cartoon - There is an electric energy associated with the position of a charge.
•Opening Demo •Warm-up problems
•Physlet
•Topics
•Electric potential energy and electric potential
•Calculation of potential from field
•Potential from a point charge
•Potential due to a group of point charges, electric dipole
•Potential due to continuous charged distributions
•Calculating the filed from the potential
•Electric potential energy from a system of point charge
•Equipotential Surface
•Potential of a charged isolated conductor
•Demos
•teflon and silk
•Charge Tester, non-spherical conductor, compare charge density at
Radii
•Van de Graaff generator with pointed objects
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Potential Energy and Electric potential
• The electric force is mathematically the same as gravity so it too must be a
conservative force. We will find it useful to define a potential energy as is the
case for gravity. Recall that the change in the potential energy in moving from
one point a to point b is the negative of the work done by the electric
force.
b
• U  Ub  Ua = - W = -Work done by the electric force =   F  ds
b
• Since F  q0 E ,
a
 U =q0  E  ds and
a
• Electric Potential difference = Potential energy change/ unit charge
V 
•
U
q0
SI unit of electric potential is volt (V):
1 Volt = 1 Joule/Coulomb (1 V = 1 J/C)
V  Vb  Va    E  ds (independent of path, ds)
• Joule is too large a unit of energy when working at the atomic or molecular level,
so use the electron-volt (eV), the energy obtained when an electron moves through
a potential difference of 1 V.
1 eV = 1.6 x 10-19 J
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f
U  U f  Ui = - Work done by the electric force =
  F  ds
i
y

V 
U
q
x
V  Vf  Vi    E  ds (independent of path, ds)
Therefore, electric force is a conservative force.
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W
F
V 
    ds    E  ds
q0
q0
•The potential difference is the negative of the work done
per unit charge by an electric field on a positive unit
charge
when it moves from one point to another.
• V is a scalar not a vector. Simplifies solving problems.
•We are free to choose V to be 0 at any location. Normally
V is chosen to be 0 at the negative terminal of a battery or
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0 at infinity for a point charge.
Example of finding the potential difference in
a Uniform Field
What is the electric potential difference for a unit positive charge moving in an
uniform electric field from a to b?
E
E
d
a
x direction
b
b
b
a
a
V   E  ds   E  dx   E ( xb  xa)
V  Ed
U  qV
dV  Edx
E  dV / dx
d
U  qEd
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Example for a battery in a circuit
•
In a 9 volt battery, typically used in IC circuits, the positive terminal has a
potential 9 v higher than the negative terminal. If one micro-Coulomb of
positive charge flows through an external circuit from the positive to negative
terminal, how much has its potential energy been changed?
V 
q
U
 Vb  Va  (0  9)V
q
U  9q
 ( 9V )  1 106 C
U  9  10 6 Joules
U   9 microJoules
=  9 J
9 J
Potential energy is lower by
We also assumed that the potential at b was 0
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Example of a proton accelerated in a uniform
field
A proton is placed in an electric field of E=105 V/m and released. After going 10
cm, what is its speed?
Use conservation of energy.
a
b
E = 105 V/m
d = 10 cm
2qEd
v
m
+
2  1.6  10 19 C  10 5 Vm  0.1m
1.67  10 27 kg
V  Vb  Va  Ed
U  qV  qEd
v
U  K  0
K  U
m
v  1.4  10
s
8
K  qEd
1 2
mv  qEd
2
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What is the electric potential when moving
from one point to another in a field due to a
point charge?
V    E  dr
kq
E  2 r̂
r
f
Vf  Vi    E  dr
i
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Potential of a point charge at a distance R
f
Vf  Vi    E  dr̂

i


1
1
1 1
V f  Vi    E  dr̂  kq cos 0  2 dr  kq
 kq(  )
r
rR
 R
R
R
q
V f  Vi  0  Vi  k
R
kq
V
R
1
k
4 0
Replace R with r
1
q
V
4 0 r
eqn 25-26
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Electric potential for a positive point charge
kq
V (r) 
r
r  x 2  y2
• V is a scalar
• V is positive for positive charges, negative for negative charges.
• r is always positive.
• For many point charges, the potential at a point in space is the
simple algebraic sum (Not a vector sum)
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Electric potential due to a positive point charge
Hydrogen atom.
• What is the electric potential at a distance of 0.529 A from
the proton? 1A= 10-10 m
 8.99  109 Nm2   1.6  10 19 C
2
kq 
C

V

R
.529  10 10 m
V  27.2
r = 0.529 A
J
 27.2Volts
C
What is the electric potential energy of the electron
at that point?
U = qV= (-1.6 x 10-19 C) (27.2 V)= - 43.52 x 10-19 J
or - 27.2 eV where eV stands for electron volts.
Total energy of the electron in the ground state of hydrogen is - 13.6 eV
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Also U= 2E = -27.2 eV. This agrees with above formula.
What is the electric potential due to several
point charges?
•
For many point charges, the potential at a point in space is the simple algebraic sum
(Not a vector sum)
kqi
V 
ri
i
y
q2
q1
r1
 q1 q 2 q 3 
V  k   
 r1 r 2 r 3 
r2
r3
q3
x
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Potential due to a dipole
For two point charges, the total potential is the sum of
the potentials of each point charge.
So, Vdipole  Vtotal  Va  Vb
 q ( q) 
Vdipole  Va  Vb  k  

r
r
b 
 a
 rb  ra 
 kq

r
r
 ab 
We are interested in the regime where r>>d.
As in fig 2, ra and rb are nearly parallel. And the
difference in their length is dcos.Also because r>>d,
ra rb is approximately r2.
Vdipole
d cos
kq
r2
kp cos
r2
where p is the dipole moment.
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Potential due to a ring of charge
•
•
Direct integration. Since V is a scalar, it is easier to evaluate V than E.
Find V on the axis of a ring of total charge Q. Use the formula for a point
charge, but replace q with elemental charge dq and integrate.
Point charge
kq
V
r
For an element of charge dq, dV 
kdq
r
r is a constant as we integrate.
V



kdq
r
kdq
(z 2  R 2 )
k
(z  R )
2
2
 dq
V
k
(z  R )
This is simpler than finding E because V
is not a vector.
2
2
Q
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Potential due to a line charge
We know that for an element of charge dq
dq
the potential is
dV  k
r
For the line charge let the charge density be .
Then dq=dx
So, dV  k
dx
r
Then , dV  k
But, r  x 2  d2
dx
x 2  d2
Now, we can find the total potential V produced by the rod at point P by
integrating along the length of the rod from x=0 to x=L
L
L
V   dV   k
0
0
dx
x d
2
L
2
 k 
0
dx
x d
2
So, V  k(ln( L  L2  d2 )  ln d)
 V  k ln(x  x  d )
2
2
2
L
0
 L  L2  d 2 
Or, V  k  ln 

d


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A new method to find E if the potential is known.
If we know V, how do we find E?
V    E  ds
dV  E  ds
ˆ  ĵdy  k̂dz
ds  idx
dV  E x dx  E y dy  E z dz
dV
dx
dV
Ey  
dy
dV
Ez  
dz
Ex  
E  Ex iˆ  Ey ĵ  Ez k̂
So the x component of E is the derivative of V with respect to x, etc.
–If V = a constant, then Ex = 0. The lines or surfaces on which V
remains constant are called equipotential lines or surfaces.
–See example on next slide
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Equipotential Surfaces
• Three examples
• What is the obvious equipotential surface and
equipotential volume for an arbitrary shaped charged
conductor?
• See physlet 9.3.2 Which equipotential surfaces fit the
field lines?
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x
Blue lines are the electric field lines
Orange dotted lines represent the equipotential surfaces
a)
Electric Dipole
(ellipsoidal concentric shells)
b)
Point charge
(concentric shells)
c)
Uniform E field
E  Ex , Ey  0, Ez  0
dV
dx
V  Ex d
Ex  
V = constant in y and z
directions
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Dielectric Breakdown: Application of Gauss’s Law
If the electric field in a gas exceeds a certain value, the gas
breaks down and you get a spark or lightning bolt if the gas
is air. In dry air at STP, you get a spark when
4 V
E  3  10
cm
V = constant on surface of conductor
Radius r2
r1
1
2
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This explains why:
•
Sharp points on conductors have the highest electric fields and cause
corona discharge or sparks.
•
Pick up the most charge with charge tester from the pointy regions of the
non-spherical conductor.
•
Use non-spherical metal conductor charged with teflon rod. Show
variation of charge across surface with charge tester.
Radius R
V = constant on surface of conductor
1
+
Cloud
+
+
+
-
-
-
2
Van de Graaff
-
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How does a conductor shield the interior from an
exterior electric field?
• Start out with a uniform electric field
with no excess charge on conductor.
Electrons on surface of conductor adjust
so that:
1. E=0 inside conductor
2. Electric field lines are perpendicular
to the surface. Suppose they weren’t?
s
3. Does E =
 0 just outside the conductor
4. Is suniform over the surface?
5. Is the surface an equipotential?
6. If the surface had an excess charge, how would your answers change?
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A metal slab is put in a uniform electric field of 106 N/C
with the field perpendicular to both surfaces
.
– Show how the charges are distributed on the
conductor.
– Draw the appropriate pill boxes.
– What is the charge density on each face of
the slab?
– Apply Gauss’s Law.
qin
 E  da   0
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What is the electric potential of a uniformly charged circular disk?
We can treat the disk as a set of ring charges.
The ring of radius R’ and thickness dR’ has an
area of 2R’dR’ and it’s charge is dq = sdA =
s(2R’)dR’ where s=Q/(R2), the surface
charge density. The potential due to the charge
on this ring at point P given by
k
V
Q
2
2
( z  (R' ) )
The potential dV at a point P due to
the charged ring of radius R’ is
kdq
ks2R' dR'
dV 

( z 2  (R' )2 )
( z 2  (R' )2 )
Integrating R’ from R’=0 to R’=R
R
V
0
ks2R' dR'
( z 2  (R' )2 )
 V  2ks( z 2  R2  z)
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