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Electric Potential 2
Q V
4pe0 R
Q
4pe0 r
R
r
C
B
r
a
B
q
r
A
A
path independence
a
Previously
•
•
•
•
Work has to be done to move a charge in an electric field
A charge has potential energy in the field
Define potential energy (U) = charge (Q) x electric potential (V)
Then
V    E dl
• A positive test charge will move toward lower potential
• Define zero potential at infinity
• Potential difference between two points
Question 1
Two test charges are brought separately to
the vicinity of positive charge Q.
Q
+
Q
+
r
– charge +q is brought to pt A, a distance r
from Q.
– charge +2q is brought to pt B, a distance 2r from Q.
q
A
2r
– Compare the potential at point A (VA) to that at B (VB):
(a) VA < VB
(b) VA = VB
(c) VA > VB
2q
B
Question 1
Two test charges are brought separately to
the vicinity of positive charge Q.
Q
+
Q
+
r
– charge +q is brought to pt A, a distance r
from Q.
– charge +2q is brought to pt B, a distance 2r from Q.
q
A
2r
2q
B
– Compare the potential at point A (VA) to that at B (VB):
(a) VA < VB
(b) VA = VB
(c) VA > VB
• THE POTENTIAL IS A FUNCTION OF THE SPACE !!!!!
• The Potential does not depend on the “test” charge at all.
A positive “test charge” would move (“fall”) from point A towards
point B.
Therefore, VA > VB
In fact, since point B is twice as far from the charge as point A, we
calculate that VA = 2VB!!
Potential Difference between two points
• We want to evaluate the potential difference from A
to B in the field of a point charge q
E
What path should we choose
to evaluate the integral?
•If we choose straight line, the
integral is difficult to evaluate.
B
•
• Magnitude
r
B
q
of the field is different at
each point along the line.
• Angle between E and path is different at
each point along line.
r
A
A
Potential difference between two points
• We want to evaluate the potential difference from A
to B in the field of a point charge q
E
C
• Instead choose path A->C then C->B
B
From A to C, E is perpendicular to
the path. i.e
r
B
q
From C to B, E is parallel to the
path. i.e
r
A
A
Potential Difference between 2 points
C
• Evaluate potential difference from
A to B along path ACB.
E
B
by definition:
r
B
Evaluate the integral:
q
r
A
A
C
Independent of Path?
B
r
B
• How general is this result?
• Consider the approximation to the
straight path from A->B (white arrow) = 2
arcs (radii = r1 and r2) plus the 3
connecting radial pieces.
r
A
A
q
B
r2
• For the 2 arcs + 3 radials path:
A
r1
q
This is the same result as above!!
The straight line path is better
approximated by Increasing the number
of arcs and radial pieces.
Potential from N charges
r1
The potential from a collection of N
charges is just the algebraic sum of
the potential due to each charge
separately.

x
q1
q2
r2
r3
q3
Question 2
+5 μC
-3 μC
A
9) Two charges q1 = + 5 μC, q2 = -3μC are placed at equal distances
from the point A. What is the electric potential at point A?
a) VA < 0
b) VA = 0
c) VA > 0
Question 2
+5 μC
-3 μC
A
Two charges q1 = + 5 μC, q2 = -3μC are placed at equal distances
from the point A. What is the electric potential at point A?
a) VA < 0
b) VA = 0
c) VA > 0
A positive test charge would move to the right
Question 3
• Which of the following charge distributions produces V(x) =
0 for all points on the x-axis? (we are defining V(x)  0 at x
=)
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
(a)
+2mC
-2mC
x
-1mC
-2mC
(b)
x
-1mC
(c)
+1mC
Question 3
• Which of the following charge distributions produces V(x) =
0 for all points on the x-axis? (we are defining V(x)  0 at x
=)
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
(a)
+2mC
-2mC
x
-1mC
-2mC
(b)
x
-1mC
(c)
+1mC
The total potential at a point is the ALGEBRAIC sum of the contributions.
Therefore, to make V(x)=0 for all x, the +Q and -Q contributions must cancel.
Thus any point on the x-axis must be equidistant from +2mC and -2mC and also
from +1mC and -1mC.
This condition is met only in case (a)!
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r<a)
I
II
III
IV
uncharged
conductor
r
solid sphere
with total
charge Q
a
c
b
Calculating Electric Potentials
Calculate the potential V(r) at the point shown (r < a)
• Where do we know the potential,
and where do we need to know it?
V=0 at r =  ...
we need r < a ...
• Determine E(r) for all regions
in between these two points
uncharged
conductor
I
II
III
IV
r a
solid sphere
with total
charge Q
• Determine V for each region by integration
c
b
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r < a)
uncharged
conductor
I
II
III
IV
r a
c
b
solid sphere
with total
charge Q
• Check the sign of each potential difference V
V > 0 means we went “uphill”
V < 0 means we went “downhill”
(from the point of view
of a positive charge)
... and so on ...
Calculating Electric Potentials
III
IV
• Look at first term:
• Line integral from infinity to c
has to be positive, pushing
against a force:

dl
r a

E
1  Q 
1 Q


4pe o  r    4pe o c
c
• What’s left?
I
II
c
b
Calculating Electric Potentials
I
II
III
IV
• Look at third term:

dl
• Line integral from b to a, again has to be
positive, pushing against a force:
1  Q 
1 Q (b  a )




4pe o  r   b 4pe o
ab
a
• What’s left?
Previous slide
we have calculated this already

E
r a
c
b
Calculating Electric Potentials
I
II
III
IV
• Look at last term:

dl
• Line integral from a to r, again has to be
positive, pushing against a force.
r a

E
• But this time the force doesn’t vary the
same way, since “r ’ ” determines the
amount of source charge
Q
r3
a
3
c
b
Q
This is the charge
that is inside “r”
and sources field
• What’s left to do?
• ADD THEM ALL UP!
• Sum the potentials
r
1  Qr  
1 Q  r2 


1  2 

3 
4pe o  2a  a 4pe o 2a  a 
2
Calculating Electric Potentials
I
II
Calculate the potential V(r) at the point shown (r < a)
• Add up the terms from I, III and IV:
I
III

dl
III
IV

E
r a
c
b
IV

dl
Potential increase
from moving into
the sphere
The potential difference
from infinity to a if the
conducting shell
weren’t there
An adjustment to
account for the fact
that the conductor
is an equipotential,
V= 0 from c → b
Calculating Electric Potentials
Summary
The potential as a function of r for all 4
regions is:
I
r > c:
II
b < r < c:
III
a < r < b:
IV
r < a:
I
II
III
IV
r a
c
b
Let’s try some numbers
Q = 6m C
a = 5cm
b = 8cm
c = 10cm
I
II
III
IV
r a
I
r > c: V(r = 12cm) = 449.5 kV
II
b < r < c: V(r = 9cm) = 539.4 kV
III
a < r < b: V(r = 7cm) = 635.7 kV
IV
r < a: V(r = 3cm) = 961.2 kV
c
b
Sparks
• High electric fields can ionize nonconducting
materials (“dielectrics”)
Dielectric
Insulator
Conductor
Breakdown
• Breakdown can occur when the field is greater than
the “dielectric strength” of the material.
– E.g., in air,
Emax  3 106 N/C  3 106 V/m  30 kV/cm
What is ΔV?
d  2mm
Ex.
Vdoorknob
V  Emax  d
Vfinger
Arc discharge equalizes
the potential
 30 kV/cm• 0.2 cm
 6 kV
Note: High humidity can also bleed the charge off  reduce ΔV.
Question 3
Two charged balls are each at the same potential V. Ball 2 is
twice as large as ball 1.
r1
Ball 1
r2
Ball 2
As V is increased, which ball will induce breakdown first?
(a) Ball 1
(b) Ball 2
(c) Same Time
Question 3
Two charged balls are each at the same potential V. Ball 2 is
twice as large as ball 1.
r1
Ball 1
r2
Ball 2
As V is increased, which ball will induce breakdown first?
(a) Ball 1
Esurface
Q
k 2
r
\
(b) Ball 2
(c) Same Time
Q
r
V
Smaller r  higher E  closer to breakdown

r
V k
Esurface
Ex. V  100 kV
100 103 V
r
 0.03m  3cm
6
3 10 V/m
High Voltage Terminals must be big!
Lightning!
+
_
+
_
+
_
Collisions produce
charged particles.
The heavier
particles (-) sit near
the bottom of the
cloud; the lighter
particles (+) near
the top.
Stepped
Leader
Negatively
charged
electrons
begin
zigzagging
downward.
Attraction
As the stepped
leader nears
the ground, it
draws a
streamer of
positive charge
upward.
Flowing
Charge
As the leader
and the
streamer come
together,
powerful
electric current
begins flowing
Contact!
Intense wave of
positive charge,
a “return stroke,”
travels upward
at 108 m/s
t ~ 30ms
Factoids: V ~ 200 M volts
I ~ 40,000 amp
P ~ 1012 W
Summary
• If we know the electric field E,
allows us to calculate the potential function V everywhere
(define VA = 0 above)
• Potential due to n charges:
• Equipotential surfaces are surfaces where the
potential is constant.