Workshop: Using Visualization in Teaching Introductory E&M AAPT National Summer Meeting, Edmonton, Alberta, Canada. Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy. MIT Class: Electric Potential. 2. Potential Energy and Potential. Start with Gravity.
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Workshop: Using Visualization in Teaching Introductory E&MAAPT National Summer Meeting, Edmonton, Alberta, Canada.Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy
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Start with Gravity
Gravitational force on m due to M:
Work done by gravity moving m from A to B:
PATH
INTEGRAL
Work done by gravity moving m from A to B:
Work Done by Earth’s GravityThinking about the sign and meaning of this…
Moving from rA to rB:
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Answer: 3. Wg is negative – we do work
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
Work done by gravity moving m from A to B:
Work Near Earth’s SurfaceG roughly constant:
Wg depends only on endpoints
– not on path taken –
Conservative Force
Define gravitational potential difference:
Gravitational Potential(Joules/kilogram)That is, two particle interaction single particle effect
Consider 3 equal masses sitting in different gravitational potentials:
A) Constant, zero potential
B) Constant, nonzero potential
C) Linear potential (V x) but sitting at V = 0
Which statement is true?
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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
Units: Joules/Coulomb = Volts
Change in potential energy in moving the charged object (charge q) from A to B:
Joules
Change in potential energy in moving the charged object (charge q) from A to B:
Joules
The external work is
If the kinetic energy of the charged object does not change,
then the external work equals the change in potential energy
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Know These!
Charges CREATE Potential Landscapes
Charges CREATE Potential Landscapes
Charges FEEL Potential Landscapes
We work with DU (DV) because only changes matter
Place a positive charge in an electric field. It will accelerate from
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
Place a negative charge in an electric field. It will accelerate from
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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:
Just like gravity, moving in field direction reduces potential
Take V = 0 at r = ∞:
+q
P
PRS: Two Point ChargesThe work done in moving a positive test charge from infinity to the point P midway between two charges of magnitude +q and –q:
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+q
P
PRS Answer: Two Point Charges3. Work from to P is zero
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)
Consider the 3 point charges at left.
What total electric potential do they create at point P (assuming V = 0)
Consider the point charges you looked at earlier:
You calculated V(P). From that can you derive E(P)?
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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!
larger than that for x < 0
smaller than that for x < 0
equal to that for x < 0
I don’t know
The graph above shows a potential V as a function of x. The magnitude of the electric field for x > 0 is
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The slope is smaller for x > 0 than x < 0
Translation: The hill is steeper on the left than on the right.
Answer: 2. The magnitude of the electric field for x > 0 is smaller than that for x < 0
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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
The above shows potential V(x). Which is true?
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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.
Answer: 2. Ex > 0 is > 0 and Ex < 0 is < 0
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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?
How much energy to put two charges as pictured?
How much energy to put three charges as pictured?
Total configuration energy:
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?
All points on equipotential curve are at same potential.
Each curve represented by V(x,y) = constant
E is perpendicular to all equipotentials
Constant E field
Point Charge
Electric dipole
A point charge q creates a field and potential around it:
Use superposition for systems of charges
They are related:
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If you put a charged particle, (charge q), in a field:
To move a charged particle, (charge q), in a field
and the particle does not change its kinetic energy
then:
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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
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1
2
3
4
5
6
5
6
4
1
3
2
The circle is at +5 V relative to the plate. Which of the below is the most accurate equipotential map?
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5
Answer:
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.
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1
2
3
4
5
6
6
5
4
3
2
1
The circle is at +5 V relative to the plate. Which of the below is the most accurate electric field line map?
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64
2
Answer:
Field lines must be perpendicular to equipotential surfaces, including the conductors themselves.
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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)
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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)
A
C
B
D
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:
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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.
A
C
B
D
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)
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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)
A
C
B
D
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:
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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 there is nearly zero.
A
C
B
D
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)
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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)
A
C
D
B
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:
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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.
A
C
D
B
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)
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no net charge
a positive charge
a negative charge
I don’t know
A drop of water falls through the right can. If the can has positive charge on it, the separated water drop will have
Can
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Water Drop
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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.
+ +

+
+
+
+
+
+

Answer: 3. The drop has a negative charge
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