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PH214_2011SPRING_EXAM1_PROFSOLN_[0]

PH214_2011SPRING_EXAM1_PROFSOLN_[0] - Ph 214 Spring 2011...

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Ph 214, Spring 2011 Exam # 1 Question #1 SOLUTIONS 1. [25 pts] Avalanche breakdown. a. (15 pts) Find an expression for vector E rad at this position for all times. Be sure to completely specify the direction (in cartesian coordinates) as well as its dependence on the given parameters and t . Electromagnetic radiation is produced by accelerating point charges. We know that the radiated electric field from a single accelerated point charge is given by vector E rad ( vector r, t ) = vectora ( t ) q 4 πǫ 0 c 2 r so our first order of business will be to determine the acceleration. We are given the position as a function of time so we can easily compute vectora ( t ) = d 2 dt 2 vector z e ( t ). We obtain d 2 dt 2 ( z 0 ) ˆ z = 0 t ≤ − t 0 vectora ( t ) = d 2 dt 2 bracketleftbiggparenleftbigg 1 6 βt 3 1 2 βt 2 0 t parenrightbigg ˆ z bracketrightbigg = βt ˆ z t 0 < t < t 0 d 2 dt 2 ( z 0 ) ˆ z = 0 t t 0 [3 pts for correct expressions. Answer must include domain and correct direction.] x x z z R 2 3 2 R R 30 observer 30 vectora vectora Figure 1: Geometry of observer relative to accelerating charges. Now that we have determined vectora , we can compute the projection of vectora perpendicular to the direction of propagation. The observer is located along a radius vector lying in the x z plane and making an angle of 30 with respect to the z axis (refer to Figure 1). For acceleration in +ˆ z , | vectora | = | vectora | sin 30 = 1 2 | vectora | . The corresponding direction is given by cos 30 ˆ x + sin 30 ˆ z = 3 2 ˆ x + 1 2 ˆ z . Therefore, 1

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Ph 214, Spring 2011 Exam # 1 Question #1 SOLUTIONS 0 t ≤ − t 0 vectora ( t ) = parenleftbigg 1 2 β t parenrightbigg parenleftBigg 3 2 ˆ x + 1 2 ˆ z parenrightBigg t 0 < t < t 0 0 t t 0 [5 pts for correct magnitude and direction. Breakdown is 1 + 3 + 1 pts for the three cases.] The distance from the charge to the observer is R and therefore the time it takes for the electromagnetic radiation to reach the observer is R/c . This implies that the retarded time is given by t = t R/c . Treating all N electrons as a single charged object of total charge Ne , we can put all the pieces together to obtain 0 t ≤ − t 0 + R c vector E rad ( R, t ) = N e β 8 πǫ 0 c 2 R parenleftbigg t R c parenrightbigg parenleftBigg 3 2 ˆ x + 1 2 ˆ z parenrightBigg t 0 + R c < t < t 0 + R c 0 t t 0 + R c [7 pts broken down as 1 + 5 + 1 pts for the three cases. To receive full credit: the direction must be correct, the retarded time must be specifically included & the time limits for each of the three cases must be correct.] b. (5 pts) For this observer, make a sketch of vector E rad as a function of time. t - t 0 + R c t 0 + R c R c Ne β t 0 8 πepsilon1 0 c 2 R - Ne β t 0 8 πepsilon1 0 c 2 R vector E rad / ˆ E rad Figure 2: Plot of vector E rad ( t ).
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