1. Basic Laws Basic Definitions Important Derivations Examples + Maxwell’s Equations Fields & Potentials Energies, Currents & Power Fields, Motion & Polarizations

2. Gauss' Law Gauss' Law (a FUNDAMENTAL Law): The net electric flux through any closed surface is proportional to the charge enclosed by that surface. How to Apply?? –Useful in finding E when the physical situation has SYMMETRY –CHOOSE a closed surface such that the integral is TRIVIAL »Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface »Magnitude: surface must be chosen such that E has the same value at all points on the surface when E is perpendicular to the surface. »Therefore: bring E outside of the integral Most of our results for E fields around points, lines, planes were obtained this way.

3. Gauss’ Law: LHS: RHS: q = ALL charge inside radius r LHS: RHS: q = ALL charge inside radius r, length L LHS: RHS: q = ALL charge inside cylinder=  A Spherical Symmetry: Cylindrical Symmetry: Planar Symmetry: Use superposition of these results

4. Gauss’ Law Example Consider What is the surface charge density  L on the left side of the conducting slab? 11 22 EAEA A LL RR Two non-conducting infinite sheets with surface charge densities  1 =+3  C/m 2 and  2 = +5  C/m 2 and an uncharged conducting slab E =0 1. Use SUPERPOSITION of Electric Fields. On Far Left: = 1/2  0 ( -  1 -  2 -  L -  R ) At X : = 1/2  0 ( +  1 -  2 -  L -  R ) In Red: = 1/2  0 (  1 -  2 +  L -  R ) = 0 On Far Right: = 1/2e 0 (  1 +  2 +  L +  R ) 2. Uncharged :  L +  R = 0 ; Add  L -  R =  1 -  2 Get  L = (  1 -  2 )/2

5. Electric Potential Difference Suppose charge q 0 is moved from pt A to pt B through a region of space described by electric field. Since there will be a force on the charge due to, a certain amount of work W will have to be done to accomplish this task. We define the electric potential difference as: Since the force we have to exert must just cancel the electric force: A B q 0

6. Potential from N charges The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately. x r1r1 r2r2 r3r3 q1q1 q3q3 q2q2 

7. allows us to calculate the potential function V everywhere (keep in mind, we often define V A = 0 at some convenient place)  If we know the electric field E everywhere, allows us to calculate the electric field E everywhere If we know the potential function V everywhere, Units for Potential! 1 Joule/Coul = 1 VOLT The Bottom Line

8. Q R1R1 R2R2 R3R3.P.P A charge of Q = 4nC is placed on a solid conducting sphere of radius R 1 = 5cm and surrounded by a concentric conducting solid shell with inner radius R 2 = 20cm and outer radius R 3 = 25cm and no net charge. Now we connect the inner sphere to the outer shell with a conducting wire. What is the potential at R 1 (assuming V = 0 at infinity)? a. kQ/ R 1 b. kQ/ R 2 c. kQ/ R 3

9. C 1 = 1  F C 2 = 2  F C 3 = 3  F 12 v V = 0 b The potential along the bottom wire is defined to be zero. The voltage V b at the point labeled b is: a. 10 V b. 5.46 V c. 4.0 V d. 2.0 V e. 0.17 V Assume that all capacitors are uncharged before the circuit is assembled.

10. The north end of a magnet is very slowly brought close to a block of some unknown substance. If the block is attracted by the magnet, the substance could equally well be either a hard ferromagnet or diamagnetic. (Assume the attraction exists even after all possible induced surface currents have died away; if it helps, remember that water is diamagnetic, and liquid oxygen is paramagnetic.) NS F a. True b. False

11. An infinitely long coaxial cable consists of a solid conducting cylinder of radius R 1 = 1mm and a hollow conducting shell with inner radius R 2 =2mm and outer radius R 3 =3mm. The cable is aligned along the z axis, and centered at x=y=0. The inner conductor carries a uniformly distributed current I 1 = 2 A in the +z direction (out of the page) and the conducting shell carries a uniformly distributed current of I 2 = 5 A in the –z direction (into the page). R3R3 P x y. R 1 = 1 mm R 2 = 2 mm R 3 = 3 mm R1R1 R2R2 The magnetic field at the point P is 0 (P is not necessarily shown to scale). How far is P from the center of the cable? a. 2.00 mm b. 2.40 mm c. 2.45 mm

12. X X X X X X X 10cm coil Top View B B side view [The two ends of the wire are connected (not shown), forming a closed circuit.] BzBz time 0.2 s A single wire is wrapped into a coil with 200 turns and diameter 10 cm (see figures). The axis of the coil is aligned vertically. The coil is placed in a uniform magnetic field of magnitude 2.0 T in the upward direction. The direction of the field is suddenly reversed during a time interval of 0.2 s. Find the average emf in the coil during the reversal. a. 0.157 V b. 15.7 V c. 31.4 V

13. Non-Simple Circuits All circuits are governed by: When capacitors ( V = Q/C ) and inductors ( V = L dI/dt ) are involved, the currents are time-dependent: R II C  a b R I I  a b L

14. AC Circuits Resonance: I m max when X L = X C LCR Circuit: L C   R  ImRImR ImXLImXL ImXCImXC mm  High “Q”: High V L, V C over narrow freq. range

15. t I T L R ε I A B The current as a function of time is shown in the right-hand side figure. Assume R and L are ideal, and R> 0, L>0. Which of the following graphs best describes the voltage across the inductor, VL(t) (=VB – VA)? t VLVL T a. t VLVL T e. t VLVL T b. t VLVL T c. t VLVL T d.

16. I A · Side View I A · End-on View (current coming out of the page) In Lecture we discussed the Poynting vector as describing the direction of energy transport in electromagnetic waves. The Poynting vector applies to other situations as well when there is a flow of electromagnetic energy. Consider the following diagram, showing a capacitor in the process of being charged up (i.e., a current is flowing onto the left plate and off of the right plate). In the Side View, what is the direction of S at the point A a. down (i.e., radially inward) b. up (i.e., radially outward) c. into the paper d. out of the paper e. undefined, since |S|=0 in this situation.

17. Circular Polarization LCP = CCW x y SLOW   E TA FAST EfEf EsEs LCP z /4 I1I1 I 2 = I 1

18. 21.Light polarized along the x-axis is directed through a quarter waveplate (fast axis initially along x) and a polarizer (transmission axis also along x), as shown below. [Assume the polarizer is ideal, i.e., light polarized along the transmission axis is completely transmitted.] z TA x y EoEo Quarter Waveplate fast axis As the fast axis of the quarter waveplate is rotated in the xy plane by a small angle, the output intensity will decrease. a. True b. False