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1 Physics 208: Electricity nd Mgnetism Finl Exm, Secs My Instructor: Dr. George R. Welch, 415 Engineering-Physics, Print your nme netly: Lst nme: First nme: Sign your nme: Plese fill in your University ID number: (Not your Socil Security number) IMPORTANT Red these directions crefully: There re 8 problems totlling 200 points. Check your exm to mke sure you hve ll the pges. Work ech problem on the pge the problem is on. You my use the bck. If you need extr pges, I hve plenty up front. Indicte wht you re doing! We cnnot give full credit for merely writing down the nswer. Netness counts! I will give generous prtil credit if I cn tell tht you re on the right trck. This mens you must be net nd orgnized. Ech problem with its ssocited figure is self explntory. If you must sk question, then come to the front, being s discrete s possible so s not to disturb others. Put your nme on ech pge it is sked for. You will lose credit if you fil to print your nme on ech pge it is sked for.

2 Problem points. Two chrged prticles re seprted by distnce 2. One prticle hs chrge +Q nd the other hs chrge Q. Let n x-y coordinte system be centered directly between the two chrges, with the x-xis long the line joining the two chrges. Clculte the electric field t point A nd the electric potentil t point B shown in the Figure. Assume tht the potentil is zero infinitely fr wy. A +y B +Q Q +x Wrning! Wrning! Wrning! Electric field is vector. I wnt both components of this vector, reltive to the coordinte system given bove. Work Netly! More points for net work.

3 Problem points. A thin rod of length L crries totl chrge Q tht is uniformly distributed long its length. Consider the point P shown by n in the Figure. Point P lies long the line of the rod nd distnce from the center (where > L/2). Clculte the electric field t point P. L/2 L/2 P Q I hve to be ble to red it to grde it. Be net.

4 Problem 3(). 20 points. A lrge flt slb of nonconducting mteril crries uniform chrge per unit volume ρ. The thickness of the slb is d nd is not negligible. The Figure shows cross section through the slb. The x-xis is perpendiculr to the slb, nd we tke x = 0 to be t the center of the slb. Clculte the electric field s function of x. Tht is, clculte E(x). Note tht there is symmetry bout x = 0, mening E(x) = E( x), so you only need to worry bout x 0. However, the field is different inside the slb nd outside. Thus, you need to derive two formuls: One for E(x) where x d/2 nd nother for E(x) where x d/2. (You might wnt to check your work by verifying tht both of these formuls give the sme vlue t x = d.) Note: You MUST show your work. No credit for writing down the nswer. ρ d x = 0 x Remember, more points if you re net.

5 Problem 3(b). 5 points. For the previous problem (3) derive n expression for the electric potentil V (x) inside the slb. As reference point, tke the electric potentil to be zero t x = 0. Work netly, nd I ll grde it esier.

6 Problem points. In the circuit shown below, the bttery hs voltge V = 10 V, the resistor hs impedence R = 10 Ω, the cpcitor hs cpcitnce C = 0.1 µf = F, nd the inductor hs inductnce L = 10 µh = H. Initilly, the switch is in the center, so the bttery is not connected to nything. The chrge on the cpcitor is zero, nd there is no current flowing. C () The switch is then moved to the up position. Clculte the potentil difference cross the cpcitor t time 2 µs (tht is, t t = seconds) fter the switch is moved up. R V L (b) Exctly 2 µs fter the switch ws moved up (s in prt ()) the switch is then moved to the (fully) down position. Clculte the potentil difference cross the cpcitor t time 1 µs (tht is, t t = seconds) fter the switch is moved down. Importnt note: Do not express your nswer in terms of pproximte deciml expressions. Tht is, do not plug numbers into clcultor. Insted, express your nswer in terms of trnscendentl functions, such s ln(1), e 2, cos(3), etc. Don t forget to be net.

7 Problem points. A DC circuit is constructed with three btteries nd three resistors with vlues s shown. Clculte the current supplied by ech bttery. 10 V 10Ω 15 V 15Ω 30 V 30Ω I cn grde your work more generously if I cn red it! Be NEAT!

8 Problem points. A squre loop of wire, of side, is plced in non-uniform mgnetic field. The field points down into the pge, nd does not depend on the y-coordinte (trnsltionlly invrint long y) but depends on the x-coordinte in the following wy: B = µ 0I 2π 2 2 x 2. where the origin of the coordinte system is t the center of the squre. Squre loop of wire y Clculte the the mgnetic flux through this loop. Note: You MUST show your work. No credit for writing down the nswer. x The following integrl my be useful: dx 2 x = 1 ( ) + x 2 2 ln x ( 2 > x 2 ) Mgnetic Field B(x) (into the pge) Did you work NEATLY? Netness counts!

9 Problem points. Consider wire with shpe s shown in the Figure. There is long stright prt, then circulr prt tht bends through 90, then nother long stright prt. The rdius of curvture of the circulr prt is r, so tht the displcement between the two long stright prts is lso r. I r P r Suppose the wire is crrying current I. Clculte the mgnetic field t the point P shown in the Figure, which is t the center of the circulr prt of the wire. Orgnize your thoughts, nd work NEATLY.

10 Problem points. () (8 points) An object is plced 10 cm from sphericl convex reflecting surfce. The rdius of curvture of this surfce is 20 cm. Clculte the position of the imge of the object. Is it rel or virtul? Is it erect or inverted? 10 cm (b) (9 points) A thin lens is mde from glss with index of refrction equl to 1.5. Both surfces re convex s shown in the Figure, nd their rdii of curvture re 4 cm nd 12 cm. An object is plced 10 cm from this lens. Clculte the position of the imge of the object. Is it rel or virtul? Is it erect or inverted? 10 cm n=1.5 (c) (8 points) A object is plced 10 cm to the right of mirror described in prt (), nd 10 cm to the left of lens described in prt (b). There re two imges of this object to the right of the lens. Clculte the positions of both of these imges. 10 cm 10 cm

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