1. Still unimportable
#845325F2,9 (for lecture 4)
#845325F2,5 (for lecture 6)
No clicker scores for:
Schlossberger,Aaron,Norman
Tokarcik,Nicholas,M

2. Hand graded problem: Students who got 5 and lower are encouraged to come to my office and show on blank page (means from scratch) how to solve Exam1 hand graded problem.
Multiple choice:
Total:
Students who got 35 and lower should talk to me about finding ways to improve their grasp of material.

3. Previous Lecture Introduced Electrostatic Potential Energy (Uel)
Electric Potential (V)
Learned how to compute V for
point charge
charged sphere

4. Potential Energy
Introduced the concept of electric field E to deal with forces
Introduced electric potential V to deal with work and energy
Electric potential: electric potential energy per unit charge
Potential energy is associated with pairs of interacting objects
A single particle has no electric potential energy

5. V due to Two Particles Electric potential is scalar:
Electric potential energy of the system: q3
If we add one more charge at position C:
Comment on sign of V

6. Question 1
What is the potential in the center of uniformly charged hollow sphere? ([k=1/(4pe0)])
k*Q/R2
k*Q/R
k*Q/(4pR2)
Not enough information Q R B

7. Potential Inside a Uniformly Charged Hollow Sphere=0

8. Potential Difference with Varying Field
In general, integration path may be complex

9. Sign of the Potential Difference
The potential difference V can be positive or negative.
The sign determines whether a particular charged particle will gain or lose energy in moving from one place to another.
If qV < 0 – then potential energy decreases and K increases
If qV > 0 – then potential energy increases and K decreases
If there are no external forces acting on the system:

10. Example x
An electron traveling to the right enters capacitor through a small hole
at A. Electric field strength is 2x103 N/C. What is the change in the
electron’s potential energy in traveling from A to B? What is its
change in kinetic energy? D(AB)= 4mm
The system is comprised of the charged capacitor plates and the electron. this example, the electric force is in the direction opposite to the displacement through which the force acts. Remember that e is a number. Also remember that the F in this expression is that due to sources within the system (F_internal).
= (1.6x10-19 C)(2x103 N/C)(0.004m) =1.3x10-18 J
DK = -DUelectric = -1.3x10-18 J

11. Example
300

12. Question 1 A proton is free to move from right to left
in the diagram shown. There are no other
forces acting on the proton. As the proton
moves from right to left, its potential energy:
Is constant during the motion
Decreases
Increases
Not enough information
V1 < V2

13. Sign of the Potential Difference
If freed, a positive charge will move to the area with a lower potential:
Vf – Vi < 0
(no external forces)
So if the potential energy decreases, K must increase since there are no external forces in this problem.
V1 < V2
Moving in the direction of E means that potential is decreasing

14. Sign of the Potential Difference
To move a positive charge to the area with higher potential:
Vf – Vi > 0
Need external force to perform work
What can we say about the proton’s kinetic energy?
V1 < V2
Moving opposite to E means that potential is increasing

15. Question 2
A system consists of a proton inside of a capacitor. The proton moves from left to right as shown at a constant speed due to the action of an external agent.
Which of the following statements are true?
V1 < V2
The proton’s potential energy is unchanged and the external agent does no work on the system.
The proton’s potential energy decreases and the external agent does work W > 0 on the system.
The proton’s potential energy decreases and the external agent does work W < 0 on the system.
The proton’s potential energy increases and the external agent does work W < 0 on the system.
The proton’s potential energy increases and the external agent does work W > 0 on the system. E.

16. Potential in Metal In static equilibrium
A Capacitor with large plates and a small gap of 3 mm has a potential difference of 6 Volts from one plate to the other. Find E E
d =3 mm +Q -Q
-3 V
+3 V
Charges are on surface
V = 6 Volt

17. Shifting the Zero Potential
In most cases we are interested in V, not the absolute values of V

18. Potential Difference in a Nonuniform Fieldx
Note: E2x is a negative number!
From A to C: DV1 = -E1x(xC-xA);
From C to B: DV2 = -E2x(xB-xC);
So, A to B: DV = DV1+ DV2 = -E1x(xC-xA) - E2x(xB-xC)

19. Example: Two Different Paths in Capacitor
Need to find VAC =VC - VA
1. Straight path AC
Electric field is constant inside capacitor!

20. Example: Two Different Paths in Capacitor
Need to find VAC =VC -VA
1. Straight path AC
2. Path AB C
Does it make sense VCB=0?
No work is required to move at right angle to the electric field

21. Example: Different Paths near Point Charge
1. Along straight radial path: rf ri +q
For final r greater than initial r, the change in potential is less than zero as expected since the path direction is the same as that of the electric field. Likewise, if we go from a larger r to a small r, the potential increases.

22. Example: Different Paths near Point Charge
2. Special case
iA:
AB:
BC: +
Cf:

23. Example: Different Paths near Point Charge
3. Arbitrary path +

24. Potential Difference: Path Independence
Path independence principle:
V between two points does not depend on integration path

25. Physics of lightning
Free path of e- in air at 1 atm, room T is about 1 micron (1*10-6m)
Ionization potential of O2 is 12.5eV, N2 15 eV
(1eV=1.6*10-19 J is a kinetic energy which e- gains by going through V=1 Volt)
Approximate that about 10V is needed to ionize air.
So, what will be E= V/x = 10/10-6 = 107 V/m
(close to E critical = 3*106 N/C we used in Lecture 7)

26. Electron-Volt (eV) – Unit of Energy
What is the change in electric potential energy associated with moving an electron from 1Å to 2Å from a proton?
If an electron moves through a potential difference of 1 V there is a change in electric potential energy of 1 eV.
1 eV = e.(1 V) = ( C)(1 V) = 1.610-19 J