1. Lecture 7-1 Electric Potential Energy of a Charge (continued) i is “the” reference point. Choice of reference point (or point of zero potential energy) is arbitrary. 0 i is often chosen to be infinitely far away (∞)

2. Lecture 7-2 Electric Potential So U(r)/q 0 is independent of q 0, allowing us to introduce electric potential V independent of q 0. [Electric potential] = [energy]/[charge] SI units: J/C = V (volts) U(r) of a test charge q 0 in electric field generated by other source charges is proportional to q 0. taking the same reference point Potential energy difference when 1 C of charge is moved between points of potential difference 1 V Scalar!

3. Lecture 7-3 Electric Potential Energy and Electric Potential positive charge High U (potential energy) Low U negative charge High U Low UHigh V (potential) Low V Electric field direction High V Low V Electric field direction

4. Lecture 7-4 E from V Expressed as a vector, E is the negative gradient of V We can obtain the electric field E from the potential V by inverting the integral that computes V from E : E always points from high V to low V. (But high V is high U only for a positive charge.)

5. Lecture 7-5 READING QUIZ 1 A conducting sphere A with radius R A = 1 meter is connected via a conducting wire to a conducting sphere B with radius R B = ½ meter. The spheres have a total charge Q = Q A + Q B. The radial electric fields at surface of the spheres are E A and E B respectively. Which of the following statements is correct? A| E A > E B B| E A = E B C| E A = 1/2 E B D| E A = 2 E B

6. Lecture 7-6 Equipotential Surfaces An equipotential surface is a surface on which the potential is the same everywhere. For a displacement Δr of a test charge q 0 along an equipotentital, U = constant on an equipotential surface. E is everywhere perpendictular to U Potential difference between nearby equipotentials is approximately equal to E times the separation distance. Equipotential surfaces are drawn at constant intervals of ΔV http://www.its.caltech.edu/~phys1/java/phys1/EField/EField.html

7. Lecture 7-7 Potential and Conductors Entire conductor including its surface(s) has uniform V. Equipotential surfaces Draw equipotential surfaces outside the conductor on which V is uniform. For Δl on equipotential, or equipotential surface no work done

8. Lecture 7-8 Potential Differences of a Uniformly Charged Sheet Electric field is uniform on each side of the sheet as shown. Equipotential surfaces are perpendictular to the electric fields. Separation between equipotential surfaces are equal to the potential differences divided by the magnitude of electric field.

9. Lecture 7-9 Reference Point for Potential of Uniformly Charged Sheet Take a reference point at O. Or take it at some other point so that we choose V(0)=V 0 : P x O equipotential since V=V(x)

10. Lecture 7-10 Two Ways to Calculate Potential Integrate - E from the reference point at (∞) to the point (P) of observation: Integrate dV (contribution to V(r) from each infinitesimal source charge dq) over all source charges:  A line integral (which could be tricky to do)  If E is known and simple and a simple path can be used, it may be reduced to a simple, ordinary 1D integral. q1q1 q2q2 q3q3 q4q4 P Q P r P Q

11. Lecture 7-11 Two Ways to Calculate Potential Integrate - E from the reference point at (∞) to the point (P) of observation: NOTE: One case where this choice of reference point fails is the infinite sheet of charge – which of course is unphysical. The E field in this idealized case goes on forever, and you get an infinity if you integrate this constant E in from a reference voltage of 0 at infinity. Of course, if you choose the reference point at the sheet of charge or near to it, you can certainly do the above integral. Q P r

12. Lecture 7-12 Potential from uniformly charged spherical shell Potential r > R: r < R: Electric field (Gauss’s Law) r < R: E = 0 r > R: E = kQ/r 2 0 (or a charged solid spherical conductor) + + + + + +

13. Lecture 7-13 Electric Field of a Uniformly Charged Sphere (non-conducting) Apply Gauss’s Law directly or use superposition of the shell results

14. Lecture 7-14 Potential of a Uniformly Charged (solid) Sphere (2) r < R (1) r > R ∞ reference same as shell or conducting sphere) (very different from conducting sphere!) V r R + + + + + + + insulator

15. Lecture 7-15 Physics 241 –warm-up Quiz 2 A infinite plane with uniform charge density +σ. What is the potential difference V B -V A ? AB 1 m 2 m +σ.+σ. a)-(3/2)σ/ε 0 b)-σ/2ε 0 c)3σ/ε 0 d)σ/2ε 0 e)-3σ/ε 0

16. Lecture 7-16 Charged Concentric Spherical Conductors Q out originally, then add Q in (a) r > c a b c Q in Q out r V (b) b < r < c (c) a < r < b (d) r < a

17. Lecture 7-17 Potential from continuous charge distribution: ring At point P on axis of ring Sum scalar contributions dV vector

18. Lecture 7-18 Plot Potential and Field of a Ring At point P on axis of ring vector x V(x) x ExEx

19. Lecture 7-19 High Electric Field at Sharp Tips Two conducting spheres are connected by a long conducting wire. The total charge on them is Q = Q 1 +Q 2. Potential is the same: The smaller the radius of curvature, the larger the electric field.

20. Lecture 7-20 DOCCAM 2 Corona Discharge Wind 5A-23

21. Lecture 7-21 Lightning rod Air “Break down” before too much charge accumulated, i.e. much weaker lightning which is much less destructive. Golf court

22. Lecture 7-22 Physics 241 10:30 Quiz 3 September 13, 2011 A spherical shell of radius 50 cm is charged uniformly with a total charge of Q Coulombs. What is the potential difference V B -V A ? a)kQ/2 b)-2kQ c)-(3/2)kQ d)kQ e)-kQ/2 Q AB 1 m 2 m

23. Lecture 7-23 Physics 241 11:30 Quiz 3 September 15, 2011 Two parallel planes, 2 m apart, are charged with uniform charge densities 2σ and -σ (in C/m 2 ). Respectively as shown. What is the potential difference V B -V A (in volts)? a)-(3/2)σ/ε 0 b)-2σ/ε 0 c) 3σ/ε 0 d) σ/ε 0 e)-3σ/ε 0 AB 2 m

24. Lecture 7-24 Physics 241 –Quiz 3 11:30 February 1. 2011 A spherical shell of radius 20 cm is charged uniformly with a total charge of Q Coulombs. What is the potential difference V B -V A ? a)kQ/2 b)-2kQ c)-(3/2)kQ d)kQ e)-kQ/2 Q AB 1 m 2 m