Download Electric Potential Powerpoint

Workshop: Using Visualization
in Teaching Introductory E&M
AAPT National Summer Meeting, Edmonton, Alberta,
Canada.
Organizers: John Belcher, Peter Dourmashkin,
Carolann Koleci, Sahana Murthy
P04 - 1
MIT Class:
Electric Potential
P04 - 2
Potential Energy
and Potential
Start with Gravity
P04 - 3
Gravity: Force and Work
Gravitational force on m due to M:
Mm
Fg  G 2 rˆ
r
Work done by gravity moving m from A to B:
B
Wg   Fg  d s
A
PATH
INTEGRAL
P04 - 4
Work Done by Earth’s Gravity
Work done by gravity moving m from A to B:
ds
W  F d s
g

B
g

GMm 

    2 rˆ   dr rˆ  rd θˆ
r

A

rB
GMm 
GMm

   2 dr 
r
 r  r
rA
 1 1 
 GMm   
 rB rA 
rB
A
P04 - 5
PRS Question:
Sign of Wg
P04 - 6
PRS: Sign of Wg
Thinking about the sign and
meaning of this…
 1 1 
Wg  GMm   
 rB rA 
Moving from rA to rB:
1.
100% 2.
0% 3.
0% 4.
0% 5.
0%
Wg is positive – we do work
Wg is positive – gravity does work
Wg is negative – we do work
Wg is negative – gravity does work
I don’t know
P04 - 7
PRS Answer: Sign of Wg
Answer: 3. Wg is negative
– we do work
 1 1 
Wg  GMm   
 rB rA 
Wg is the work that gravity does. This is the opposite
of the work that we must do in order to move an object
in a gravitational field.
We are pushing against gravity  we do positive work
P04 - 8
Work Near Earth’s Surface
GM
G roughly constant: g   2 yˆ   g yˆ
rE
Work done by gravity moving m from A to B:
mg yˆ   d s

A
Wg   Fg  d s  
B
yB
   mgdy  mg ( yB  y A )
yA
Wg depends only on endpoints
– not on path taken –
Conservative Force
P04 - 9
Potential Energy (Joules)
B
 U g  U B  U A    Fg  d s   Wg
A
GMm
GMm
(1) Fg  
rˆ  U g  
 U0
2
r
r
(2) Fg  mg yˆ
 U g  mgy  U 0
• U0: constant depending on reference point
• Only potential difference U has
physical significance
P04 - 10
Gravitational Potential
(Joules/kilogram)
Define gravitational potential difference:
 Vg 
U g
m
B
B
A
A
   (Fg / m)  d s    g  d s
Just as Fg  g ,  U g   Vg
Force
Field
Energy
Potential
That is, two particle interaction  single particle effect
P04 - 11
PRS Question:
Masses in Potentials
P04 - 12
PRS: Masses in Potentials
Consider 3 equal masses sitting in different
gravitational potentials:
A)
Constant, zero potential
B)
Constant, non-zero potential
C)
Linear potential (V  x) but sitting at V = 0
Which statement is true?
20%
20%
20%
20%
20%
0%
1.
2.
3.
4.
5.
6.
None of the masses accelerate
Only B accelerates
Only C accelerates
All masses accelerate, B has largest acceleration
All masses accelerate, C has largest acceleration
I don’t know
P04 - 13
PRS Answer: Masses in Potentials
Answer: 3. Only C (linear potential) accelerates
When you think about potential, think “height.”
For example, near the Earth:
U = mgh so V = gh
Constant potential (think constant height) does
not cause acceleration!
The value of the potential (height) is irrelevant.
Only the slope matters
P04 - 14
Move to Electrostatics
P04 - 15
Gravity - Electrostatics
Mass M
M
g  G 2 rˆ
r
Fg  mg
Charge q (±)
q
E  ke 2 rˆ
r
FE  qE
Both forces are conservative, so…
B
 Vg    g  d s
A
B
 U g    Fg  d s
A
B
 V   E  d s
A
B
 U    FE  d s
A
P04 - 16
Potential & Potential Energy
B
 V   E  d s
Units:
Joules/Coulomb
A
= Volts
Change in potential energy in moving the
charged object (charge q) from A to B:
U  U B  U A  qV
Joules
P04 - 17
Potential & External Work
Change in potential energy in moving the
charged object (charge q) from A to B:
U  U B  U A  qV
Joules
The external work is
Wext   K   U
If the kinetic energy of the charged object
does not change,  K  0
then the external work equals the change
in potential energy
Wext   U  q V
P04 - 18
How Big is a Volt?
•
•
•
•
•
•
AA, C, D Batteries
Car Battery
US Outlet
Residential Power Line
Our Van de Graaf
Big Tesla Coil
1.5 V
12 V
120 V (AC)
P04 - 19
Potential: Summary Thus Far
Charges CREATE Potential Landscapes
r
V (r )  V0   V  V"0" 
E

d
s

"0"
P04 - 20
Potential Landscape
Positive Charge
Negative Charge
P04 - 21
Potential: Summary Thus Far
Charges CREATE Potential Landscapes
r
E

d
s

V (r )  V0   V  V"0" 
"0"
Charges FEEL Potential Landscapes
U  r   qV  r 
We work with U (V) because
only changes matter
P04 - 22
2 PRS Questions:
Potential & Potential Energy
P04 - 23
PRS: Positive Charge
Place a positive charge in an electric field. It
will accelerate from
25%
25%
25%
25%
1. higher to lower electric potential;
lower to higher potential energy
2. higher to lower electric potential;
higher to lower potential energy
3. lower to higher electric potential;
lower to higher potential energy
4. lower to higher electric potential;
higher to lower potential energy
P04 - 24
PRS Answer: Positive Charge
Answer:
2. + acc. from higher to lower electric potential;
higher to lower potential energy
Objects always “move” (accelerate) to
reduce their potential energy. Positive
charges do this by accelerating
towards a lower potential
U  q  V
P04 - 25
PRS: Negative Charge
Place a negative charge in an electric field. It
will accelerate from
20%
20%
20%
40%
1. higher to lower electric potential;
lower to higher potential energy
2. higher to lower electric potential;
higher to lower potential energy
3. lower to higher electric potential;
lower to higher potential energy
4. lower to higher electric potential;
higher to lower potential energy
P04 - 26
PRS Answer: Negative Charge
Answer:
4. Neg. acc. from lower to higher electric potential
higher to lower potential energy
Objects always “move” (accelerate) to
reduce their potential energy. Negative
charges do this by accelerating
towards a higher potential:
U  q  V
P04 - 27
Potential Landscape
Positive Charge
Negative Charge
P04 - 28
Creating Potentials:
Calculating from E,
Two Examples
P04 - 29
Potential in a Uniform Field
B
V  VB  VA    E  d s
A
TO
FROM
   Eˆj  ds  E  dy
B
B
A
A
  Ed
Just like gravity, moving in field
direction reduces potential
E   Eˆj
ds  dyˆj
P04 - 30
Potential Created by Pt Charge
B
V  VB  VA   E  ds
A
B dr
rˆ
   kQ 2  d s  kQ  2
A
A r
r
1 1
 kQ   
 rB rA 
B
Take V = 0 at r = ∞:
kQ
VPointCharge (r ) 
r

rˆ
E  kQ 2
r

d s  dr rˆ  r d θˆ
P04 - 31
PRS Question:
Point Charge Potential
P04 - 32
PRS: Two Point Charges
The work done in moving a positive test charge
from infinity to the point P midway between two
charges of magnitude +q and –q:
+q
20% 1.
20% 2.
20% 3.
20% 4.
20% 5.
P
-q
is positive.
is negative.
is zero.
can not be determined – not enough info is given.
I don’t know
P04 - 33
PRS Answer: Two Point Charges
3. Work from  to P is zero
+q
P
-q
The potential at  is zero.
The potential at P is zero because equal and
opposite potentials are superimposed from the
two point charges (remember: V is a scalar,
not a vector)
P04 - 34
Potential Landscape
Positive Charge
Negative Charge
P04 - 35
Group Problem: Superposition
Consider the 3 point
charges at left.
What total electric
potential do they
create at point P
(assuming V = 0)
P04 - 36
Deriving E from V
P04 - 37
B
Deriving E from V
 V   E  d s
A
A = (x,y,z), B=(x+x,y,z)
 s   x ˆi
( x  x , y , z )
V  

E  d s  E   s  E  ( x ˆi )   Ex  x
( x, y, z )
V
 V Ex = Rate of change in V
Ex  

x
 x with y and z held constant
P04 - 38
Deriving E from V
If we do all coordinates:
 V ˆ V ˆ V ˆ 
E  
i+
j
k
y
z 
 x
  ˆ  ˆ  ˆ
 
i+
j  k V
y
z 
 x
E   V
Gradient (del) operator:
 ˆ  ˆ  ˆ

i
j+ k
x
y
z
P04 - 39
PRS Questions:
E from V
P04 - 40
PRS: E from V
Consider the point charges you looked at earlier:
V  P    kQ a
You calculated V(P). From that can you derive E(P)?
20% 1.
20% 2.
20% 3.
20% 4.
20% 5.
Yes, its kQ/a2 (up)
Yes, its kQ/a2 (down)
Yes in theory, but I don’t know how to take a gradient
No, you can’t get E(P) from V(P)
P04 - 41
I don’t know
PRS Answer: E from V
4. No, you can’t get E(P) from V(P)
The electric field is the gradient (spatial
derivative) of the potential. Knowing the
potential at a single point tells you nothing
about its derivative.
People commonly make the mistake of
trying to do this. Don’t!
P04 - 42
PRS: E from V
The graph above shows a potential V as a function
of x. The magnitude of the electric field for x > 0 is
0%
1. larger than that for x < 0
0%
2. smaller than that for x < 0
0%
3. equal to that for x < 0
:20P04 0%
4. I don’t know
43
PRS Answer: E from V
Answer: 2. The magnitude of the electric field for
x > 0 is smaller than that for x < 0
The slope is smaller for x > 0 than x < 0
Translation: The hill is steeper on the
left than on the right.
P04 - 44
PRS: E from V
The above shows potential V(x). Which is true?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
Ex > 0 is > 0 and Ex < 0 is > 0
Ex > 0 is > 0 and Ex < 0 is < 0
Ex > 0 is < 0 and Ex < 0 is < 0
Ex > 0 is < 0 and Ex < 0 is > 0
I don’t know
20
P04 - 45
PRS Answer: E from V
Answer: 2. Ex > 0 is > 0 and Ex < 0 is < 0
E is the negative slope of the potential,
negative on the left, positive on the right
Translation: “Downhill” is to the left on the left
and to the right on the right.
P04 - 46
Potential (V)
Group Problem: E from V
10
5
0
-5
0
5
Z Position (mm)
A potential V(x,y,z) is plotted above. It does
not depend on x or y.
What is the electric field everywhere?
Are there charges anywhere? What sign?
P04 - 47
Demonstration:
Making & Measuring
Potential
(Lab Preview)
P04 - 48
Configuration Energy
P04 - 49
Configuration Energy
How much energy to put two charges as pictured?
1) First charge is free
2) Second charge sees first:
U12  W2  q2V1 
1
q1q2
4 o r12
P04 - 50
Configuration Energy
How much energy to put three charges as pictured?
1) Know how to do first two
2) Bring in third:
q3  q1 q2 

W3  q3  V1  V2  


4 0  r13 r23 
Total configuration energy:
1  q1q2 q1q3 q2 q3
U  W2  W3 



4 0  r12
r13
r23

  U12  U13  U 23

P04 - 51
Group Problem: Build It
1) How much energy
did it take to assemble
the charges at left?
2) How much energy
would it take to add a
4th charge +3Q at P?
P04 - 52
Equipotentials
P04 - 53
Topographic Maps
P04 - 54
Equipotential Curves
All points on equipotential curve are at same potential.
Each curve represented by V(x,y) = constant
P04 - 55
Direction of Electric Field E
E is perpendicular to all equipotentials
Constant E field
Point Charge
Electric dipole
P04 - 56
Properties of Equipotentials
• E field lines point from high to low potential
• E field lines perpendicular to equipotentials
• Have no component along equipotential
• No work to move along equipotential
P04 - 57
Summary: E Field and Potential:
Creating
A point charge q creates a field and potential around it:
q
q
E  ke 2 rˆ ; V  ke
r
r
Use superposition for
systems of charges
They are related:
B
E   V ;  V  VB  VA   E  d s
A
P04 - 58
E Field and Potential: Effects
If you put a charged particle, (charge q), in a field:
F  qE
To move a charged particle, (charge q), in a field
and the particle does not change its kinetic energy
then:
Wext  U  qV
P04 - 59
Experiment 1: Equipotentials
Download LabView file (save to desktop)
and run it
Log in to server and add each student to
your group (enter your MIT ID)
Each group will do two of the four figures
(your choice). We will break about half
way through for some PRS
P04 - 60
PRS Questions:
Midpoint Check
P04 - 61
PRS: Lab Midpoint: Equipotential
The circle is at +5 V relative to the plate. Which of the below is
the most accurate equipotential map?
:20
4
1
0%
0%
0%
0%
6
0%
5
0%
4
6
3
3
6
2
5
1
2
3
4
5
6.
1
2
1.
2.
3.
4.
5.
P04 - 62
PRS Answer: Equipotential
Answer:
5
The electric field is stronger between the plate and
circle than on either outer side, so the equipotential
lines must be spaced most closely in between the two
conductors.
P04 - 63
PRS: Lab Midpoint: Field Lines
20
The circle is at +5 V relative to the plate. Which of the below is
the most accurate electric field line map?
6
0%
0%
0%
0%
0%
0%
6
3
5
5
4
2
3
4
1
1
1
2
3
4
5
6
2
1.
2.
3.
4.
5.
6.
P04 - 64
PRS Answer: Field Lines
Answer:
2
Field lines must be perpendicular to
equipotential surfaces, including the
conductors themselves.
P04 - 65
Experiment 1: Equipotentials
Continue with the experiment…
If you finish early make sure that you talk
about the extra questions posed at the
end of the lab. Labs will be asked about
on the exams (see, for example, the final
exam from Fall 2005)
P04 - 66
PRS Questions:
Lab Summary
P04 - 67
PRS: Lab Summary: Potentials
Holding the red plate at +5 V relative to the ground of
the blue plate, what is true about the electric potential
at the following locations:
A
B
C
D
0%
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
6.
V(A) > V(B) > V(C) > V(D)
V(A) > V(B) ~ V(C) > V(D)
V(A) ~ V(B) > V(C) ~ V(D)
V(D) > V(C) ~ V(B) > V(A)
V(B) > V(C) > V(D) ~ V(A)
V(A) > V(D) ~ V(C) > V(B)
20
P04 - 68
PRS Answer: Potentials
Holding the red plate at +5 V relative to the ground of
the blue plate…
Answer: 2. V(A) > V(B) ~ V(C) > V(D)
A
C
B
D
The potential at A is nearly +5 V.
The potential at B & C ~ 2.5 V (they are both halfway).
The potential at D is about 0 V.
P04 - 69
PRS: Lab Summary: E Field
Holding the red plate at +5 V relative to the ground of
the blue plate, what is true about the electric field at the
following locations:
A
C
B
D
0%
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
6.
E(A) > E(B) > E(C) > E(D)
E(A) > E(B) ~ E(C) > E(D)
E(A) ~ E(B) > E(C) ~ E(D)
E(D) > E(C) ~ E(B) > E(A)
E(B) > E(C) > E(D) ~ E(A)
E(A) > E(D) ~ E(C) > E(B)
20
P04 - 70
PRS Answer: E Fields
Holding the red plate at +5 V relative to the ground of
the blue plate…
Answer: 5. E(B) > E(C) > E(D) ~ E(A)
A
C
B
D
The potential changes most rapidly (and hence E is
largest) at B. It also changes at C, but not as fast.
The potential is very uniform outside, so the E field out
P04 - 71
there is nearly zero.
PRS: Lab Summary: Charge
Holding the red plate at +5 V relative to the ground of
the blue plate, what is true about the amount of charge
near the following points:
A
C
D
B
0%
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
6.
|Q(A)| ~ |Q(C)| > |Q(B)| ~ |Q(D)|
|Q(A)| > |Q(B)| ~ |Q(C)| > |Q(D)|
|Q(A)| ~ |Q(B)| > |Q(C)| ~ |Q(D)|
|Q(D)| ~ |Q(C)| > |Q(B)| ~ |Q(A)|
|Q(B)| ~ |Q(D)| > |Q(A)| ~ |Q(C)|
|Q(A)| > |Q(D)| ~ |Q(C)| > |Q(B)|
20
P04 - 72
PRS Answer: Charge
Holding the red plate at +5 V relative to the ground of
the blue plate…
Answer: 3. |Q(A)| ~ |Q(B)| > |Q(C)| ~ |Q(D)|
A
C
D
B
Charges go where the field is highest (higher
field  more field lines  more charges to
source & sink). Field at A & B is the same, so
Q is as well. Higher than at C & D.
P04 - 73
PRS: Kelvin Water Dropper
20
A drop of water falls through the right can. If the
can has positive charge on it, the separated water
drop will have
0%
0%
0%
0%
1.
2.
3.
4.
no net charge
a positive charge
a negative charge
I don’t know
Can
Water Drop
P04 - 74
PRS Answer: Kelvin Water Dropper
Answer: 3. The drop has a negative charge
The positive charge on
the can repels positive
charge to the top of the
drop and attracts
negative charge to the
bottom of the drop just
before it separates.
After the drop separates
its charge is therefore
negative.
+
+
+
+
+
-
-
+
+
+
P04 - 75