Material covered: Chapters

Transcription

1 Lecture 6. Electrostatic Potential The first common hour midterm eam will be held on Thursday October 5, 9:50 to 11:10 PM (at night) on the Busch campus. You should go to the room corresponding to the first 3 letters of your last name. If you have a conflict with the eam time, please contact Prof. Montalvo with your entire schedule for the week of October 2 at your earliest convenience but not later than 5:00 pm on Wednesday, September 27. Aaa Hal ARC 103 Haq Mou Hill 114 Mug Seh PHY LH Sen Zzz SEC 111 Material covered: Chapters

2 Lectures 7-18 (Sept. 28 Nov. 9) will be given by Prof. Vitaly Podzorov. I ll be away for Sept. 28 Oct. 23 (no office hours, no contact), please contact Profs. Podzorov and Montalvo with all your questions.

3 Electrostatic Potential Scalar field! Electrostatic potential V r = U r q Units: J/C=Volt = Potential is related (but not equal!) to the potential energy of charges in the electric field. Numerically, this is the work done by eternal forces ( us ) to bring a positive unit charge from some reference point to the point in question. Potential of a positive charge V 6

4 From El. Potential to El. Field F, y = U, y U, y y y E, y = V, y V, y y y F, y = qe, y U, y = qq, y Potential of a positive charge V, y = ccccc E, y =? E, y = ccccc 7

5 Calculation of Electrostatic Potential q 2 1. Discrete charges V r = k i q i r r i V r r V = 0 q 1 r 1 O V r 2. Continuous charge distribution V r = k voo ρ r r r V = 0 dd r r O 3. E r is known V r = E r dl r ree.ppppp 8

6 Eample: Electric Field and Potential of a Charged Ring Find V() along the ais of symmetry. III Which one should we use? I II E I V = 1 4πε 0 Q 2 + a 2 II V = 1 4πε 0 λ r r r dd = 1 4πε 0 λλλ 2 + a 2 = 1 4πε 0 Q 2 + a 2 III V = 0 Q 4πε a a = y = a 2, dd = 2 = Q 1 4πε a 2 dd y 3/2 9

7 Electric Field and Potential of a Charged Metal Sphere q=q net r - which one should we use? Though the (continuous) charge distribution is given, it is easier to use our result for the field: V r r>r = E r dl r r = q dd 4πε 0 r 2 = r = q r 4πε 0 r 2 ddr q 4πε 0 1 r 1 = q 4πε 0 r V r r<r =? V r r<r = q 4πε 0 R 12

8 Electric Field and Potential of a Parallel-Plate Capacitor Potential is a continuous function of r. (eception: point charges). E E E V V V rrr rrr 15

9 Equipotential Lines and Surfaces lines (surfaces) of constant potential, plotted for fied potential increments. +5V 0V -5V -10V -15V At each point, a field line is perpendicular to an equipotential surface: aaaaa eeeeeee. sssssss E r dl = 0 16

10 Equipotential Surfaces (cont d) A B C 19

11 Conducting Surface = Equipotential Surface W=0! F E dl = 0 V=2V V=1V V=0V V=-1V V=-2V The surface of a conductor is always an equipotential surface: by moving a test charge along the surface, we don t do any work (the force on the charge is normal to its trajectory). E=0 Since E=0 inside, any point in the volume of a conductor is at the same potential as its surface. 20

12 Conclusion Electrostatic potential energy and electrostatic potential. Connection between V(r) and E(r) Equipotential surfaces and field lines. Net time: Lecture 7. Capacitors and Electric Field Energy 24.1,

13 Eample: Dipole V = +2V V = +1V V = 0V V = 1V 22

14 Intersecting Equipotential Lines If an equipotential line crosses itself, then E=0 at this point. Recall: a field line never crosses itself! 23

15 Dangers of the Web Compare these two plots. What s wrong with the right one? The electric field intensity is proportional to the density of equipotential lines. At the center E, y, z = V,y,z Touching equipotential lines (corresponding to different potentials) imply an infinitely strong field.. 24

16 Appendi I. Dangers of the Web (cont d) Four equal charges as a square there are five equilibrium points (the green circles); only the one in the center of the square is stable. A test charge placed near this point would invariably end up in the equilibrium position, which is a local minimum for the potential. The other four points are local maima of the potentials. What s wrong with these field lines/potential surfaces? It s all wrong: the field lines cross at the center (where no charges eist), and the potential forms a local minimum in the space free of charges. 25

17 Appendi II. Earnshaw s Theorem A B C D The plots show equipotential lines that correspond to configurations of point charges. Which plot is incorrect? Earnshaw's theorem: A charged body cannot be held in stable stationary equilibrium by electrostatic forces from other charged bodies. For a system governed by electrostatics there can be no potential energy minimum, or maimum, in an unoccupied by charges region of space. At a local energy minimum a positively charged particle would have to feel a restoring force no matter which way it is displaced. This could only be at a point where lines of force all converge. Likewise, the only stable position for a negative charge would be where lines of force all diverge. But each of these conditions is true only for a point where there is a particle of the opposite charge. By the same token, the only local energy maimum for a particle is where there is another particle of the same charge. The Earnshaw's Theorem eplains why you are not accustomed to seeing things levitating by electrostatic forces. For the same reason, we cannot use electrostatic forces for plasma confinement in thermonuclear power plants. 26

18 Potentials and Scalar Fields Variations of U g due to inhomogeneous g(,y) Gravitational potential energy U g, y = mg, y h y V g, y y Gravitational potential V g, y = g, y h J kk Electric field vector field. Electrostatic potential scalar field. It is often easier to find a solution for the scalar potential and then restore the electric field. Temperature Field 29