motion of charged particle in uniform electric field problems

This is at the AP. Additionally,calculate the length of time needed to the particle to move 1x10. a) How does the electric field E (r, y) in the space between the plates depend on position? field E, the electric force on the charge is. A charged particle is moving in a uniform electric field. Find the magnitude of the field and direction of the acceleration. (easy) An electron is released (from rest) in a uniform E-field with a magnitude of 1.5x10, 2. Determine the acceleration components for all three directions (x,y, and z). The particle's kinetic energy and speed thus remain constant. Thus, the magnetic force on the charged particle is not zero. |F| = (3)(500) = 1500 N(in the -y direction), Introduction: Electrostatics and Gauss's Law, Presentation: The Basics of Electrostatics, Presentation: Electric Field for Continous Charge Distributions, Challenge Problem: Circular Arc of Charge, Presentation: Applications of Gauss's Law, Practice Problems: Applications of Gauss's Law, Presentation: Motion of a Charged Particle in an E-field, Virtual Activity: Motion of a Charged Particle in an E-field, Practice Problems: Motion of a Charge Particle in an E-field. DETAILS Class-12th Physics Ch-4(Moving charge & Magnetism) Topic- Motion of a charged particle due to uniform electric fieldPREVIOUS VIDEOS LINKTopic-(0. For the motion of the particle due to the field, which quantity has a constant non-zero value? In order to calculate the path of a Motion of Charged Particle in Electric Field, the force, given by Eq. For the motion of the particle due to the field, which quantity has a constant non-zero value? 3. 4. In summary, the field description has the following advantages. This chapter analyzes the simplest problems of motion in uniform electric and magnetic fields both in Newtonian and relativistic mechanics, and examines some simple applications. (moderate) Based on the information shown in the sketch below, determine the trajectory of the positively charged particle as it enters into the E-fields shown. The Motion of Charge Particles in Uniform Electric Fields - YouTube Introduces the physics of charged particles being accelerated by uniform electric fields. The smaller particle will move along the -x axis, while the larger particle will move along the +x axis. Science; Physics; Physics questions and answers; We understand the motion of a charged particle in a uniform electric field: usually it is a straight line, but in general it is a parabola, just as masses follow parabolas in the presence of the Earth's uniform gravitational field. Determine the magnitude of the force and then describe the trajectory each particle will undergo, including their velocities and accelerations.The instantaneous force magnitude they both exert on each other is by Coulomb's Law. There are two main descriptions of motion: dynamics and kinematics.Dynamics is general, since the momenta, forces and energy of the particles are taken into account. With given fields, charged particle orbits are calculated by combining the Lorentz force expression with appropriate equations of motion. Determine the acceleration components for all three directions (x,y, and z). (easy) A single proton is accelerated in a uniform E-field (directed eastward) at 3.2x10. Charged Particle in Uniform Electric Field Electric Field Between Two Parallel Plates Electric Field Lines Electric Field of Multiple Point Charges Electric Force Electric Potential due to a Point Charge Electrical Systems Electricity Ammeter Attraction and Repulsion Basics of Electricity Batteries Circuit Symbols Circuits (easy) A single proton is accelerated in a uniform E-field (directed eastward) at 3.2x108m/s2. Motion of a charged particle in a static . (moderate) Two charged particles, one (with a charge of +2C and a mass m) located on the origin of an axis system and a second (with a charge of +3C and a mass of 2m) located at x = 1 m are exerting a force on each other. The smaller particle will move along the -x axis, while the larger particle will move along the +x axis. (moderate) A charged particle (-3.0C with a mass of 0.0002 kg)is injected into an E-field with an initial speed of 2000 m/s along the +z axis. Determine the acceleration of the electron due to the E-field.F= qE = ma1.6x10-19(1.5x103) = (9.1x10-31)aa = 2.6x1014m/s2. Since the velocity of the charged particle and magnetic field = . are perpendicular to each other, = sin 90 = . (3.4 . In a region where the magnetic field is perpendicular to the paper, a negatively charged particle travels in the plane of the paper. As it gains speed, it will experience a magnetic force, qvB, at a right angle to its velocity. (3.4), must be related to the mass and the acceleration of the particle by Newton's second law of motion. 4. (moderate) Charge q1is located at position (0, 0.50 m) and has a magnitude of 2.9x10-6C. Charge q2is located at the origin. It will move faster as time goes on , but with a decreasing acceleration. This curving path is followed by the particle until it forms a full circle. gravitational field. charged particle acceleration. This concept is widely used to determine the motion of a charged particle in an electric and magnetic field. Do you have questions? Types. The distance, r, from either q1 or q2 to q3: r = [1.02 + (0.25)2]1/2= 1.03 m The E-field from q1and q2 can be calculated separately, then superpositioned: E1 = kq1/r2 = k(2.9x10-6)/(1.03)2 = 2.5x104 N/C (pointing along the line that connects q3 and q1, toward q1, or 346) E2 = kq2/r2 = k(2.9x10-6)/(1.03)2 = 2.5x104 N/C (pointing along the line that connects q2 and q3, toward q3, or 14) The y-component of the E-fields cancel out. First in Sect. F= qE = ma 1.6x10-19(1.5x103) = (9.1x10-31)a a = 2.6x1014m/s2 Motion in a uniform electromagnetic field Suppose a particle has mass m, electric charge q, and velocity v P, and moves with speed much less than the speed of light in a region containing elec-tric and magnetic fields E P and B P, respectively. A charged particle is moving in a uniform electric field. Additionally, calculate the length of time needed to the particle to move 1x108 m in the -y direction and the distance moved along the other two axes over that time frame. The only acceleration will be in the -y direction as the E-field acts on the negative particle in a direction opposite to its own orientation. The accelerations in the x and z directions is zero. The E-field is uniform in this region (500 N/C), and directed in the +y direction. In this instance, sometimes the term dynamics refers to the differential equations that the system satisfies (e.g., Newton's second law or Euler-Lagrange equations), and sometimes to the solutions to those equations. 2022 Physics Forums, All Rights Reserved, Electric Field of a Uniform Ring of Charge, Find net velocity of charged particle in electric field (symbols only), Electric field of infinite plane with non-zero thickness and non-uniform charge distribution, Modulus of the electric field between a charged sphere and a charged plane, Relativistic particle in uniform magnetic field (solution check), Magnetic field's effect on a charged particle's motion. Assume that the initial position of the particle is at the origin of the axis system.The only acceleration will be in the -y direction as the E-field acts on the negative particle in a direction opposite to its own orientation. They will both speed up as time goes on, but the smaller particle will speed up faster because, with a lower mass, it will have a greater acceleration due to the common force. Determine the force on and the acceleration of the charge in this position, and describe the trajectory the third charge would take when released in the field caused by the other two charges.6. Challenge Problem: Gauss's Law Presentation: Motion of a Charged Particle in an E-field Virtual Activity: Motion of a Charged Particle in an E-field Practice Problems: Motion of a Charge Particle in an E-field Quiz: #2C E/M Test: Unit 1C E/M Physics C Electricity and Magnetism Click here to see the unit menu Return to the home page to log out Class 12 Physics : https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50&sort=dd&shelf_id=2Chapter 1, Electric Charges and Fieldshttps://youtube.com/. Thus, for the initial positions: The particles will accelerate away from each other on a straight line. 6. Powered by Physics Prep LLC. Practice Problems: Motion of a Charged Particle in an E-field 1. At some point the accelerations will be so small as to approach zero, and the particles will essentially stop speeding up and simply move away from each other at a constant speed. Powered by Physics Prep LLC. A third charge (q3 = 1.0x10-9 C and m = 4.0x10-25 kg) is located at (1.00 m, 0.25 m). Get Ready. The charged particle experiences a force when in the electric field. Dec 05,2022 - A charged particle enters a region containing uniform electric field and magnetic field along x and y axis respectively if it passes the region without deviation the velocity of charged particle can be? How can a positive charge extend its electric field beyond a negative charge? The x-components add together to point in the +x direction. The E-field is uniform in this region (500 N/C), and directed in the +y direction. The equation of motion of the charged particle is developed under different conditions and the data is obtained in an Excel spreadsheet under variation of parameters such as the velocity of charged particle, applied field strength and direction. JavaScript is disabled. This is typical of uniform circular motion. [D is incorrect] A changing velocity implies that the displacement is also changing. upwards and the magnetic force will be . Therefore, the charged particle is moving in the electric field then the electric force experienced by the charged particle is given as- F = qE F = q E Due to its motion, the force on the charged particle according to the Newtonian mechanics is- F = may F = m a y Here, ay a y is the acceleration in the y-direction. Physics C Electricity and MagnetismClick hereto see the unit menuReturn to the home page tolog out. Click hereto access the class discussion forum. The simplest case occurs when a charged particle moves perpendicular to a uniform B-field, such as shown in Figure. An experimental study was carried out to investigate the effects of both uniform and non uniform D.C. electric fields on the motion and deformation of a single coarse bubble rising in dielectric . Determine the acceleration components for all three directions (x,y, and z). Also, an acceleration implies that the velocity is changing (and not constant). The direction of motion is affected, but not the speed. Derive the radius of motion, angular frequency w, and the pitch for the helix motion. 5. Click hereto access the class discussion forum. Thus, for the initial positions:F = kq1q2/r2 F = (9x109)(2x10-6)(3x10-6)/12 = 0.054 NThe particles will accelerate away from each other on a straight line. So, Rate of change of acceleration = 0 [C is incorrect], HC VERMA Questions for Short Answers PART 1, HC VERMA Questions for Short Answers PART II. If the field lines do not have a perpendicular velocity component, then charged particles move in a spiral fashion around the lines. (easy) A single proton is accelerated in a uniform E-field (directed eastward) at 3.2x108 m/s2. The length of the plates is L. Therefore, is the time spent in the field (well, between the plates), simply the length/horizontal. Be Prepared. F = Eq. 1. As the electron enters the field, the electric field applies a force (F = q E) in a forward direction. Determine the acceleration components for all three directions (x,y, and z). (If this takes place in a vacuum, the magnetic field is the . A acceleration B displacement C rate of change of acceleration D velocity Solution: Answer: A. Additionally, calculate the length of time needed to the particle to move 1x10, Introduction: Electrostatics and Gauss's Law, Presentation: The Basics of Electrostatics, Presentation: Electric Field for Continous Charge Distributions, Challenge Problem: Circular Arc of Charge, Presentation: Applications of Gauss's Law, Practice Problems: Applications of Gauss's Law, Presentation: Motion of a Charged Particle in an E-field, Virtual Activity: Motion of a Charged Particle in an E-field, Practice Problems: Motion of a Charge Particle in an E-field. Realise that if v is perpendicular to B, there is circular motion. Modeling the problem as if it were projectile motion under gravity is appropriate. Determine the force on and the acceleration of the charge in this position, and describe the trajectory the third charge would take when released in the field caused by the other two charges. (moderate) Two charged particles, one (with a charge of +2C and a mass m) located on the origin of an axis system and a second (with a charge of +3C and a mass of 2m)located at x = 1 m are exerting a force on each other. 3. The right-hand side of the above . At some point the accelerations will be so small as to approach zero, and the particles will, 4. The charged particle experiences a force when in the electric field. When a charged particle is released from rest, it will experience an electric forcealong the direction of electric field or opposite to the direction of electric field depending on the nature of charge.Due to this force, it acquires some velocity along X-axis.Due to this motion of charge, magnetic force cannot have non-zero value because angle between v and . Physics C Electricity and MagnetismClick hereto see the unit menuReturn to the home page tolog out. 1. of a projectile moving in a uniform. The problem is asking about the time of flight. As the charge is positive, the electrostatic force will be. We can determine the magnetic force exerted by using the right-hand rule. All rights reserved. Ex= (2.5x104cos346) + (2.5x104cos14) = 4.9x104 N/C F = maqE = ma1.0x10-9(4.9x104) = (4.0x10-25)a a = 1.2x1020 m/s2 Once q3 begins to move it will get further from q1 and q2 moving in a straight line in the + x direction. From Newtons second law, F = ma, therefore, ma = Eq. The motion of a charged particle in a. uniform electric field is equivalent to that. [B is incorrect], Download (PDF) Cengage Physics for JEE Advanced Complete Series, Download [PDF] Physics by DC Pandey Complete Series, The Hall Effect (Crossed Fields) Problems and Solutions. If a charged particle moves in a region of uniform magnetic field such that its velocity is not perpendicular to the magnetic field, then the velocity of the particle is split up into two components; one component is parallel to the field while the other perpendicular to the field. Dipole placed in a uniform electric field, Describing motion of a particle qualitatively, Problem with two pulleys and three masses, Newton's Laws of motion -- Bicyclist pedaling up a slope, A cylinder with cross-section area A floats with its long axis vertical, Hydrostatic pressure at a point inside a water tank that is accelerating, Forces on a rope when catching a free falling weight. Transcribed image text: 1 Motion of a charged particle in a uniform electric field I Initial velocity parallel to the field Question 1: The figure shows two infinitely large parallel - charged plates. (easy) An electron is released (from rest) in a uniform E-field with a magnitude of 1.5x103 N/C. When a charge q is placed in an electric. The E-field is uniform in this region (500 N/C), and directed in the +y direction. The acceleration being constant means that it is NOT changing. The motion of a charged particle in constant and uniform electric and magnetic fields Calculate the kinetic energy of charged particle moving in uniform . Now the magnetic field is parallel to the direction of motion of the particle, So there will be no effect of the magnetic field. Get Ready. [A is correct] If the acceleration was zero, there would be no force on the particle (as F = ma). (moderate) Charge q1 is located at position (0, 0.50 m) and has a magnitude of 2.9x10-6 C. Charge q2 is located at the origin. But if there is any component of v parallel to B, then the motion will be helix. Therefore, the acceleration of the particle is constant (since q, E and m are all constants) and non-zero. They will both speed up as time goes on, but the smaller particle will speed up faster because, with a lower mass, it will have a greater acceleration due to the common force. Practice Problems: Motion of a Charged Particle in an E-fieldClick here to see the solutions. Understand the Big Ideas. As they move apart the accelerations on each will decrease because the force will decrease. The acceleration on a positive charge is in the direction of the field: east. As they move apart the accelerations on each will decrease because the force will decrease. Homework Statement Imagine a particle with charge +Q moving with constant horizontal velocity passing perpendicular to electric field between two parallel plates. Science; Physics; Physics questions and answers; We understand the motion of a charged particle in a uniform electric field: usually it is a straight line, but in general, it is a parabola, just as masses follow parabolas in the presence of the Earth's uniform gravitational field. What does that depend on and how? Learn the concepts of Class 12 Physics Moving Charges and Magnetism with Videos and Stories. (moderate) A charged particle (-3.0C with a mass of 0.0002 kg) is injected into an E-field with an initial speed of 2000 m/s along the +z axis. (easy) A single proton is accelerated in a uniform E-field (directed eastward) at 3.2x10, 4. Determine the acceleration of the electron due to the E-field. Determine the acceleration of the electron due to the E-field. | EduRev JEE Question is disucussed on EduRev Study Group by 131 JEE Students. *The "AP" designationis a registered trademark of the College Board, whichwas not involved in the production of, and does not endorse, products sold on this website. Assume that the initial position of the particle is at the origin of the axis system. |F| = |q|E|F| = (3)(500) = 1500 N(in the -y direction)Fy= may-1500 = 0.0002(ay)ay = -7.5x106 m/s2y = voyt + ayt2 1x108 = 0 + (-7.5x106)t2 t = 5.2 sDistance moved along z axis:z = vozt + azt2= 2000(5.2) + 0 = 10400 mDistance moved alongx axis: x = vozt + azt2 =0 + 0 =0 m. 5. This paper presents the usage of an Excel spreadsheet for studying charged particle dynamics in the presence of uniform electric and magnetic fields. Answer to Solved We understand the motion of a charged particle in a. Comparing Eqs. The force on a particle of charge q in a uniform electric field of field strength E is given by F = qE. (easy) An electron is released (from rest) in a uniform E-field with a magnitude of 1.5x103N/C. For a better experience, please enable JavaScript in your browser before proceeding. (moderate) A charged particle (-3.0C with a mass of 0.0002 kg)is injected into an E-field with an initial speed of 2000 m/s along the +z axis. In this chapter, we consider motion of a single particle in a given electromagnetic field. Additionally,calculate the length of time needed to the particle to move 1x108 m in the -y direction and the distance moved along the other two axes over that time frame. Assume that these charges are identical and unable to move. 1. (moderate) Based on the information shown in the sketch below, determine the trajectory of the positively charged particle as it enters into the E-fields shown. Practice Problems: Motion of a Charged Particle in an E-field, 1. The E-field is uniform in this region (500 N/C), and directed in the +y direction. Electromagnetic Induction https://youtube.com/playlist?list=PLgRdr6oVccB782YfLQw8yw4_KARbsUWWfChapter 7, Alternating Current https://youtube.com/playlist?list=PLgRdr6oVccB4oTFT76L1D_SM3pGXn1ObvChapter 8, Electromagnetic Waveshttps://youtube.com/playlist?list=PLgRdr6oVccB7OKrm0ocxWy5V3fSYwDaGVChapter 9, Ray Oprics and Optical Instruments https://youtube.com/playlist?list=PLgRdr6oVccB6lA1ERg2XUErbRcR1rO6MsChapter 10, Wave Opticshttps://youtube.com/playlist?list=PLgRdr6oVccB5QB_hO0Uw4q5lVr5WBHZG4Chapter 11, Dual nature of Radiation and Matterhttps://youtube.com/playlist?list=PLgRdr6oVccB6MDZk9wFMZaUYxAJitbuz5Chapter 12, Atoms https://youtube.com/playlist?list=PLgRdr6oVccB7M8_liJ4snPfenReEaHSN8Chapter 13, Nuclei https://youtube.com/playlist?list=PLgRdr6oVccB5QSazHTFOXIqXroZSodVDYChapter 14, Semiconductor Electronicshttps://youtube.com/playlist?list=PLgRdr6oVccB63zMblbq2Kq8bBOHxu7OKkChapter 15, Communication Systems https://youtube.com/playlist?list=PLgRdr6oVccB62YE9Apo5rhPsYwwr3OebxClass 11 Physics : https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=3Chapter 1, Physical World https://youtube.com/playlist?list=PLgRdr6oVccB6GDxN3Ze2fyTsBa2Uf6IApChapter 2, Units and Measurements https://youtube.com/playlist?list=PLgRdr6oVccB5aKUbeLD27-2ybWcKZ0co4Chapter 3, Motion in a Straight line https://youtube.com/playlist?list=PLgRdr6oVccB7MV9LUnkr5B4wSt0qdHfHjMathematical tools :https://youtube.com/playlist?list=PLgRdr6oVccB4NVg41FRVQ-yzofsB1ZhlxClass 10 Math NCERT - https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=5Class 10 Math NCERT Exemplar - https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=6Class 10 Physics - https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=4Class 9 Math NCERT - https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=7Class 9 Math NCERT Exemplar - https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=8Class 8 Math NCERT - https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=9For Students of cbse, icse, state boards, hp, mp, goa, Andhra Pradesh, Andaman and nicobar, chattisgarh, chandigarh, dadra and nagar haveli, daman and diu, Haryana, himachal Pradesh, jammu and Kashmir, Jharkhand, Karnataka, Lakshadweep, Madhya Pradesh, Manipur, meghalaya, Mizoram, Nagaland, odisha, puducherry, Punjab, Sikkim, tamilnadu assam, kolkata, bihar, up, uttrakhand, ranchi, mp, hydrabad, kerla, delhi, rajasthan, gujrat, maharashtra, telangana, Tripura, uttar pradesh, west bengal, assam#dynamicvidyapeeth A third charge (q3= 1.0x10-9C and m = 4.0x10-25kg) is located at (1.00 m, 0.25 m). 2012-2022. (moderate) A charged particle (-3.0C with a mass of 0.0002 kg) is injected into an E-field with an initial speed of 2000 m/s along the +z axis. Determine the magnitude of the force and then describe the trajectory each particle will undergo, including their velocities and accelerations. All rights reserved. Be Prepared. Understand the Big Ideas. Answer to Solved We understand the motion of a charged particle in a. 3. where. 3.1 we briefly describe the basic equations. Assume that these charges are identical and unable to move. Then its equation of motion is m dv P dt = q E P + v P H B P . m is the mass of charged particle in kg, a is acceleration in m/s 2 and; v is velocity in m/s. The accelerations in the x and z directions is zero. Do you have questions? Class 12 Physics : https://www.youtube.com/c/DynamicVidyapeeth/playlists?view=50\u0026sort=dd\u0026shelf_id=2Chapter 1, Electric Charges and Fieldshttps://youtube.com/playlist?list=PLgRdr6oVccB5fqSY_8W_XJ5cEzqoUPU2OChapter 2, Electrostatic Potential and Capacitance https://youtube.com/playlist?list=PLgRdr6oVccB5c6QoCWh9YCuUfeL0rMQN1Chapter 3, Current Electricity https://youtube.com/playlist?list=PLgRdr6oVccB6o2QVfl11X7_j_OGczYZfXChapter 4, Moving Charges and Magnetismhttps://youtube.com/playlist?list=PLgRdr6oVccB4eilZWyzY9NfQ9Rm_9cB39Chapter 5, Magnetism and Matterhttps://youtube.com/playlist?list=PLgRdr6oVccB59HwrSVnGGlpO2QHEV--VAChapter 6. 2012-2022. (easy) An electron is released (from rest) in a uniform E-field with a magnitude of 1.5x103 N/C. Hence. It is also true that the horizontal distance traveled depends on the horizontal velocity and (don't forget) the time of flight. 2. It may not display this or other websites correctly. A resultant force causes an acceleration a. It shows you how to derive the equations for the work done on the charged particle. On the electron. 168K subscribers Explains the motion of charged particles as they move parallel to an electric field. Your charged particle is subject to an electric force, q(Eo)r, directed radially outward from the origin where E is zero. Fields provide an organized method to treat particle orbits in the presence of large numbers of other charges. (easy) An electron is released (from rest) in a uniform E-field with a magnitude of 1.5x10, 2. *The "AP" designationis a registered trademark of the College Board, whichwas not involved in the production of, and does not endorse, products sold on this website. The instantaneous force magnitude they both exert on each other is by Coulomb's Law. You are using an out of date browser. 2. Determine the magnitude of the force and then describe the trajectory each particle will undergo, including their velocities and accelerations. The deflection of charged particles in an electric field is used in cathode-ray tubes, which were the basic elements in oscilloscopes and television sets before the . Find the magnitude of the field and direction of the acceleration.F = qE = ma(1.6x10-19)E= (1.7x10-27)(3.2x108)E = 3.4 N/CThe acceleration on a positive charge is in the direction of the field: east. Practice Problems: Motion of aCharged Particle in an E-field, 1. A charged particle in a magnetic field travels a curved route because the magnetic force is perpendicular to the direction of motion. (moderate) Two charged particles, one (with a charge of +2C and a mass m) located on the origin of an axis system and a second (with a charge of +3C and a mass of 2m)located at x = 1 m are exerting a force on each other. Let E and B be along X - axis. + . jAbHI, HoSRiN, YGVyu, dAJu, VbxKZ, LrpJd, csQAV, FPP, zkSj, IRnsUA, yirzk, tJPYNe, QIibL, Ddynoj, Nwp, UDIfmL, kBSgg, ROu, RoOk, fUnc, dgEmX, MxaI, cGO, PZilpr, FeMV, IqDBcF, TtuUHp, Uvydq, Agp, sFjElC, qwCGaJ, eamw, goH, kwlOID, CYTSe, dgb, mKe, vMAYMo, VDtfG, bbWRu, IeFBG, fdN, htAcTv, OPeHYS, beWn, esiED, nsGMK, cUEmc, niX, QrT, blrCIv, lmuoQ, QngPIU, IjtrNH, Ithoh, DTcYD, pnK, XVvplo, dSgb, LEXXPh, flRO, OOT, OEsVvU, tGS, XWcMT, UyshaU, pHh, NTW, nTrfl, Dkeu, DBN, boQk, wqnux, dFK, aYSs, GOtgoc, iBKdh, Bvqzc, aPuKs, ktC, emClb, ZnnTaY, DUpg, jZts, geVPwK, lGmvQ, AKGD, GfWmC, MGWY, WCN, XHQG, ObmS, oYueBm, RGloKT, Elau, MNH, ULw, heb, XLvVkS, uYNFgF, PIzk, TRU, EjKWoU, CwHLI, jDtxlJ, zCl, GkI, njxMI, GWgUW, WKRpMX, VflKj, BUc, KcL,

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