# Physics 111. Exam #1. January 24, 2014

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1 Phyic 111 Exam #1 January 24, 2014 Name Pleae read and follow thee intruction carefully: Read all problem carefully before attempting to olve them. Your work mut be legible, and the organization clear. You mut how all work, including correct vector notation. You will not receive full credit for correct anwer without adequate explanation. You will not receive full credit if incorrect work or explanation are mixed in with correct work. So erae or cro out anything you don t want graded. Make explanation complete but brief. Do not write a lot of proe. Include diagram. Show what goe into a calculation, not jut the final number. For example p m v = ( 5kg) ( 2 m ) =10 kg m Give tandard SI unit with your reult unle pecifically aked for a certain unit. Unle pecifically aked to derive a reult, you may tart with the formula given on the formula heet including equation correponding to the fundamental concept. Go for partial credit. If you cannot do ome portion of a problem, invent a ymbol and/or value for the quantity you can t calculate (explain that you are doing thi), and ue it to do the ret of the problem. All multiple choice quetion are worth 3 point and each free-repone part i worth 9 point Problem #1 /24 Problem #2 /27 Problem #3 /21 Total /72 I affirm that I have carried out my academic endeavor with full academic honety.

2 1. Suppoe that you are given the ytem of point charge, where q 1 = +2µ i located at the point (x,y) = (0,0.5m), while q 2 = 6µ i located at the point (x,y) = (0, 0.2m). a. What i the electric field at a point P with coordinate (x,y) = (0.7m, 0.4m)? The component of the net electric field vector are given a E net,x = E 1,x E 2,x = kq 1 r coθ kq coθ 2 2 1,P r 2,P E net,y = E 1,y + E 2,y = kq 1 r inθ + kq inθ 2 2 1,P r 2,P From the geometry of the ytem we can determine the ditance between each charge and point P along with the value of each of the trig function. coθ 1 = = 0.61 coθ = = 0.97 inθ 1 = = 0.79 inθ 2 = = 0.28 r 1,P = ( 0.7m) 2 + ( 0.9m) 2 =1.14m r 2,P = ( 0.7m) 2 + ( 0.2m) 2 = 0.72m Inerting the quantitie into the expreion for the net horizontal and vertical component of the electric field we find E net,x = Nm ( 0.61) ( 0.97) 2 ( 1.14m) 2 ( 0.72m) 2 = N E net,y = Nm m ( 0.79) ( 0.28) ( ) 2 ( 0.72m) 2 = N Therefore the net electric field at point P i given a 2 2 E net,p = E net,x + E = tan E 1 net,y E net,p = N E = 78.8o above the negative x-axi.

3 b. How much work i required to bring in a third charge q 3 = 5µ and place it at point P? kq W = q 3 ΔV P,2 q 3 ΔV P,1 = q kq 1 0 r P,2 r P,1 = kq q q 1 r P,2 r P,1 W = Nm ( ) m = 0.3J 1.14m c. If q 3 were releaed from ret at point P, it would 1. accelerate in the direction of the net electric field at point P. 2. accelerate in the direction oppoite to the net electric field at point P. 3. feel no net force and thu remain at ret at point P. 4. feel no net force and continue moving at a contant velocity along the charge original direction of motion. d. Suppoe that q 3 were placed again at point P. If all three charge had identical mae and if the charge were releaed from ret imultaneouly, when all three charge are very far away from each other their peed would be given by 1. v = kq 1Q 2 Q 3 3rm. 2. v = 3kQ Q Q rm 6 Q 3. v = 1 Q 2 + Q 2Q 3 + Q 1Q 3. m r 1,2 r 2,3 r 1,3 2 Q 4. v = 1 Q 2 + Q 2Q 3 + Q 1Q 3. 3m r 1,2 r 2,3 r 1,3 There wa no correct anwer given to thi quetion o everyone got credit for the quetion. The correct anwer hould be v = 2k Q 1 Q 2 + Q 2Q 3 + Q 1Q 3 3m r 1,2 r 2,3 r 1,3.

4 2. A proton i accelerated from ret through a potential difference of ΔV acc = 2.3MV a hown below. a. How much work (in ev and J) wa done on the proton and what i it peed when it leave the accelerating region? ( )[ 0 V acc ] = ev acc = 2.3MeV. Thi The work done i given by W = qδv = e energy convert to W = qδv = ( e) [ 0 V acc ] = ev acc = 2.3MeV. The work done i equal to the change in kinetic energy of the proton. Thu the final peed of the proton i given by W = ΔKE = 1 m v 2 2 p p v p = 2W = J m p kg = m ΔV acc The proton that leave the accelerator above and i directed vertically upward and approache a econd et of capacitor plate angled at θ = 37 o with repect to the horizontal a hown below. The proton enter thi econd et of capacitor plate through the left-mot hole in the bottom plate. The capacitor plate are ued to teer the proton by 90 o and make it leave through the right-mot hole in the bottom plate. +Q -Q θ = 37 o ΔV acc

5 b. What electric field i needed to make the proton enter through the left hole and exit though the right hole if the ditance between the center of the hole i L = 0.5mand the plate are eparated by d = 0.1m? (Hint: Since, the proton i o mall, you can aume that it enter at the center of the left hole and exit at the center of the right hole.) The econd capacitor i inclined at θ = 37 o, the proton enter the left-mot hole and it velocity vector make a φ = 90 o 37 o = 53 o angle with repect to the lower capacitor plate. Thu the horizontal and vertical component of the initial velocity are v ix = v i coφ = m co53 = m v iy = v i inφ = m in53 = m Auming that the x-axi run along the lower capacitor plate and the y-axi i perpendicular to the lower capacitor plate, we find the time that i needed for the charge to cover the ditance L along the lower capacitor plate between the hole. The time i given from the horizontal trajectory equation where the horizontal acceleration i zero. Thu, x f = x i + v ix t t = L 0.5m = = v ix m Then to calculate the electric field that i needed we ue the vertical trajectory equation. Thu we have y f = y i + v iy t + 1 a t 2 2 a 0 = v iy + 1 a t t = 0 ( 2 y )t ( v iy + 1 a t. The acceleration i 2 y ) = 0 given by a y = F y = ee. ombining thee two reult we can olve for the m p m p electric field. We have 0 = v iy + 1 a t ( 2 y ) = v iy ee 2m t v ee iy p 2m L p v ix E = 2m v v p iy ix = kg m m = N el m and the direction of the field i from the upper plate to the lower plate. c. What i the charge Q on either plate, if the plate each have an area A = 0.1m 2 and the pace between the plate i filled with air with dielectric contant κ =1? The charge i given by Q = V = ( Ed) = κε oaed = κε o AE d Q = m N = = 7.9µ Nm 2

6 3. The Earth atmophere can act a a capacitor, with the ground acting a one plate, the cloud acting a the econd plate and the pace between the cloud and ground filled with air. Air i normally an inulating material, but under certain condition can be made to conduct electricity, o that electric charge can flow from the cloud to the ground, in what we call a lightning trike. Aume that the cloud are uniformly ditributed around the entire Earth at a fixed ditance of 5000m (~ 3mi)above the 2 ground of area 4πR Earth, where R Earth = 6400km. Further, aume that the air between the cloud and the ground ha a reitance taken to be R = 350Ω. a. Taking the upper negative plate to be the cloud and the lower poitive plate to be the ground, what i the magnitude of the difference in potential that exit between the cloud and the ground if in a typical day a maximum of of charge i pread over the urface of the Earth? The capacitance i given a: = κε 0 A d ( ) 2 = π m Nm m = 0.91F. The magnitude of the difference in potential that exit between the cloud and the ground i Q = V V = Q = F = V b. Approximately how long would it take the Earth-cloud capacitor to dicharge of 99%it total initial charge, Q max? Further, auming that the charge i immediately replenihed a oon a the dicharge proce end, how many lightning trike are produced in a ingle day if each trike contain 25 of charge? The time to dicharge 99%it total initial charge, Q max i given a t R Q t t = Rln 0.01 ( ) = 0.01Q max = Q max e ( ) = 300Ω 0.91F ln( 0.01) =1467 In thi time we dicharge of charge by lightning trike that contain of charge each. Thu we have trike trike ~ 14. onverting thi to 25 lightning trike per day we have 14 trike hr 1hr 1day =1.2 trike 106 day.

7 c. Of coure if you ve ever driven in the rain (and epecially during a thundertorm) you probably have had the occaion to et your car windhield wiper to wipe the window at interval that match the amount of rainfall hitting the window. Your car ha intermittent windhield wiper that control when the wiper actually move acro the window and you can elect how quick or low thi occur by uing a R circuit. A charging R circuit control the intermittent windhield wiper in your car by uing the car battery, which i rated at V B. Suppoe that for a particular etting, the wiper are triggered when the voltage acro a capacitor reache V, where V V B. At thi point the capacitor i quickly dicharged (through a much maller reitor) and the cycle repeat. What variable reitance R hould be ued in a charging circuit if the wiper are to operate once every t econd, where t i the amount of time between each wipe cycle of the wiper? t R = ln 1 V V B t R = ln 1 V B V t R = ln 1 V B V t R = ln 1 V V B Jut for reference, the charging/dicharging circuit i given below. When charging the capacitor, the witch S i connected to the battery (in the left mot poition), capacitor and variable reitor R (the one with the arrow through it). When the potential reache a pecified value, the witch S move to the right mot poition and dicharge the capacitor through the wiper and the wiper motor move the actual wiper. After the dicharge the witch move back to the left and the circuit charge the capacitor again. The witch move back and forth at the time t above. Thi circuit hould look reaonably familiar and not all of the control circuitry i hown. R S V B Wiper

8 Electric Force, Field and Potential Electric ircuit Light a a Wave F = k Q Q 1 2 r ˆ E = F q r 2 E Q = k Q r r ˆ 2 PE = k Q 1Q 2 r V ( r) = k Q r E x = ΔV Δx W = qδv Phyic 111 Equation Sheet Magnetic Force and Field Light a a Particle & Relativity Nuclear Phyic ontant g = 9.8 m 2 1e = k = 1 = Nm 2 4πε 2 o ε o = eV = J µ o = 4π 10 7 Tm A c = m h = J Nm 2 m e = kg = 0.511MeV c 2 m p = kg = 937.1MeV c 2 m n = kg = 948.3MeV c 2 1amu = kg = 931.5MeV c 2 N A = Ax 2 + Bx + = 0 x = B ± B2 4A 2A Geometry ircle : = 2πr = πd A = πr 2 Triangle : A = 1 bh 2 Sphere : A = 4πr 2 V = 4 3 πr3 Mic. Phyic 110 Formulae F = Δ p Δt = Δ( mv) = ma Δt F = ky F = m v 2 R r ˆ W = ΔKE = 1 m v f v i PE gravity = mgy PE pring = 1 2 ky 2 x f = x i + v ix t a xt 2 v fx = v ix + a x t v 2 fx = v 2 ix + 2a x Δx ( ) = ΔPE

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