Discussion Question 7A P212, Week 7 RC Circuits

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Eunice Harper
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1 Discussion Question 7A P1, Week 7 RC Circuits The circuit shown initially has the acitor uncharged, and the switch connected to neither terminal. At time t = 0, the switch is thrown to position a. C a E b R1 R E = 1 V C = 5 F R 1 = 3 R = 6 (a) At t = 0+, immediately after the switch is thrown to position a, what are the currents I 1 and I across the two resistors? What does the uncharged acitor look like to the rest of the circuit at time 0? Does it offer any resistance to the flow of charge? (Why or why not?) At t = 0, the acitor acts like wire w/ no resistance to current flow. Hence the battery is effectively hooked to R 6 I / R 1 V / 6 A (b) After a very long time, what is the instantaneous power P dissipated in the circuit? After a very long time, what will have happened to the acitor? Now what will it look like to the rest of the circuit? After a long time the acitor acts like an open circuit and the battery current flows through R R. Hence I = /( R R.) = 1/9 =4/3 A and the power is 1 1 P I 1V 4/3A 16 W (c) After a very long time, what is the Q charge on the acitor? To determine Q, you need the voltage across the acitor... From part (b), we know after a long time I=4/3 A. The voltage drop across the acitor is the same as the voltage drop across R V = IR =4/3 A 34 V. Q CV = 5 F 4 V =0 C 1 1

2 Next, after a very long time T, the switch is thrown to position b. a C E b R 1 R (d) What is the time constant that describes the discharging of the acitor? We have a nice formula available for time constants: = RC. But the R in the formula refers to the total resistance through which the acitor discharges. Redrawing your circuit might help you to determine this R. When the switch is thrown to b, the acitor discharges through R and R in parallel. 1 R R R / R + R 3(6)/(36) ; CR 5F 10s equiv 1 1 equiv (e) Write down an equation for the time dependence of the charge on the acitor, for times t > T. Your answer for Q(t) should depend only on the known quantities E, R 1, R, C, and T. You know the general form for the time dependence of a discharging acitor. All you have to do is fix the constants in this expression to match the charge at t = T and at t =. We know that Q will involve a exp(-t/ factor added to a possible constant. The boundary conditions are at t T, Q 0 C and fades to zero at infinity. t T The form that does this is Q 0Cexp 10s (f) What is the charge Q 0 on the acitor 0 sec after time T? t T 0s Q C C 10s 10s 0 exp 0 exp 0e.71 C (g) What is the current through R 0 sec after time T? The voltage across R is the V Q / C.71 C/ 5F 0.54 V 0 V I R I 0.54 V / A

3 Discussion Question 7B P1, Week 7 Lorentz Force on Moving Charges The figure below depicts a region of space (shaded) where there is a constant magnetic field, B = 1. T, directed along the negative z axis (into the page). The magnetic field region extends infinitely far in the positive and negative y directions, but is constrained to the region between x = 0 and x = 1 m. A proton travels along the x axis toward this region, with initial speed v 0. proton: m = 1.67 x 10-7 kg q = e = 1.6 x C v 0 = x 10 8 m/s y B = 1. T x q, m 1 m (a) What is the radius of curvature of the proton's path after it enters the magnetic field region? To describe the particle's circular path, match the Lorentz force exerted by the B field to the centripetal force. The Lorentz force provides the centripetal force to allow the proton to accelerate about a circle Using F ma in a direction perpendicular to the proton velocity: F qv B F qvb v mv 1.67e 7 kge8 m/ s F qvb ma m r 1.74 m r qb 1.e-19 C 1. (b) At what x and y positions will the proton leave the magnetic field region? Sketch the proton's path on the figure, and use your knowledge of the radius of curvature. 1 From figure x 1 m and R sin sin (1/1.74) 0.61 rad. y R-Rcos cos 0.61 rad 0.316m

4 proton: m = 1.67 x 10-7 kg q = e = 1.6 x C v 0 = x 10 8 m/s y B = 1. T x q, m 1 m (c) What is the magnitude v of the proton s speed as it exits the magnetic field? If the speed of the proton has changed, what has happened to its kinetic energy? Can the Lorentz force cause such an effect? Because the Lorentz force is always to the proton path, it does 8 no work no change in speed thus v v0 10 m/s (d) Suppose the proton had an initial component of velocity in the +z direction of v z = 1 x 10 8 m/s, and the same initial speed v 0 as before in the xy plane. What then would be its final velocity v as it exits the magnetic field? This time, express your answer as a vector with three components. Analyze this situation by components. What effect does the Lorentz force have on the particle's motion in the z direction? What about in the xy-plane? The Lorentz force is in the x-y plane hence no z acceleration. 8 Thus vz is unchanged vz =110 m/s. From diagram you can v v x y v 8 8 0cos 1 10 cos(0.61) m/s v 8 8 0sin 1 10 sin(0.61) m/s (e) Finally, a constant electric field E is added in the shaded region. The effect of this field is that all charged particles launched with initial velocity v 0 in the x direction travel straight through the field region without being deflected at all! What is the magnitude and direction of In order that the proton is undeflected, we need the electrical force to cancel the centripetal E? ˆ force due to the magnetic force. The magnetic force is initially along the +y axis. We thus want the electric force to be along the -y ˆ axis. To balance the forces we want: 8 8 F qe F qb v E B v 1. T 10 m/s E.410 y ˆ N/C E B z 0 z 0

5 Discussion Question 7C P1, Week 7 RC Circuits The circuit shown initially has the acitor uncharged, and the switch open. At time t = 0, the switch is thrown. Write all answers in terms of E, R, and C as needed. E C R I R (a) At t 0, immediately after the switch is thrown what is the current I(0 )? Consider applying KVL around the loop that includes the battery and the acitor. V IR V IR I R At t=0 Q =0 0 0 (0 ) / (b) After a very long time, what is the current I? When the acitor is fully charged it acts like an open circuit. Hence all current flows through R R. The current is then I = / R. 1 (c) After a very long time, what is the charge on the acitor Q? Consider the KVL that just consists of the acitor and the one resistor. IRV 0 IR- Q / C 0 R- Q / C 0Q C R (d) Assume Qt () Q (1exp( t gives the charge on the acitor as a function of time. Use dq 0 dq I dt 0 dt 0 What does your imply for the effective resistance that you use in R? and your answers to (a) and (c) to compute. Why does I 0 effective C?

6 Initially there is no current flow through the resistor in parallel with the acitor Hence I(0) flows entirely through the acitor. The rate of change of charge dq dq d t is the current hence 0 I 0. 1 dt Q e Q dt dt 0 0 dq C RC R Q I0 Reffective dt 0 R RC Essentially the two resistors act as if they were in parallel during the charging operation. We would get the same time constant if we discharged the acitor by shorting the battery/ (e) Use KVL for a loop that includes the acitor and the battery to compute I() t. Check that your answers are consistent with your answers to parts (a) and (b). Q 1 Q 1 1 C t V IR 0 IR 0I I 1exp C R C R C RC 1 1 t t I 1 exp 1exp R RC R RC t At t 0,1exp I which agrees with (a) RC R t As t,1exp 1 I which agrees with (b) RC R

Circuits Capacitance of a parallel-plate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = cross-sectional area) Resistance : R = ρ L / A

Circuits Capacitance of a parallel-plate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = cross-sectional area) Resistance : R = ρ L / A k = 9.0 x 109 N m2 / C2 e = 1.60 x 10-19 C ε o = 8.85 x 10-12 C2 / N m2 Coulomb s law: F = k q Q / r2 (unlike charges attract, like charges repel) Electric field from a point charge : E = k q / r2 ( towards