The charge to mass ratio of the electron was first measured by J. J. Thompson in 1897. This is oftenreferred to as THE discovery of the electron. Thompson used a combination of electric and magneticfields in his experiment. We will defer the magnetic part of the experiment to later in the course,and examine the electric part now. The device he used sent a beam of electrons between two plates,as shown below, between two plates. The initial velocity, vo of the beam was horizontally to theright. Thompson then measured the deflection, y, of the beam from where it exited the plates withthe electric field turned off. Thompson also used a magentic field, which allowed him to determinethe initial velocity. For now, lets find an expression for the charge to mass ratio in terms of thisvelocity and the other parameters.yJEPlatePlatem, Qx=Lx(a) Assume that the weight of the electron can be neglected, so that the only force in play is theelectric force. Derive an expression (in terms of the given variables E, L, y, and vo) for thisratio of the electron charge to its mass, m/Q showing and explaining all your steps. Hint: thisis very similar to a kinematics/free fall problem, two dimensional motion under the influenceof a constant force, with the constant acceleration a determined by the electric force instead ofgravity. Additional hints are provided in the homework guide at the end of this document.(b) Look up the charge and mass of the electron and design an experiment (choose values for E,L, and vo that give a reasonable deflection y.) We do want to make sure the electric force ismuch larger than the gravitational force on the electron. Note that electron velocities can bevery close to the speed of light, since the electron mass is so very small.

(a) As the electron enters the region of electric field which is in the vertical direction, it…

Answered: 0 Plate Plate m,Q x=L x (a) Assume that…

Similar questions

Consider a particle of charge q = 2.4 C and mass m = 1.5 kg passing through the region between a pair of infinitely long horizontal plates separated by a distance d = 4.5 m with a uniform electric field strength E = 36 N/C directed in the downwards direction (-y direction). The particle begins moving horizontally with an initial velocity of v = 25 m/s from a position halfway between the plates. A.How far horizontally in meters will the particle travel before striking one of the plates. B.Caculate the speed in meters per second,with which the particle will strike the plate. C.Suppose that the eletric field is directed upward instead of downward.Caculate the new horizontal distance, in meters, that the particle travels before striking one the plates.
A simple and common technique for accelerating electrons is shown in the figure, which depicts a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through. Randomized Variables E = 2.7 × 104 N/C Calculate the horizontal component of the electron's acceleration if the field strength is 2.7 × 104 N/C. Express your answer in meters per second squared, and assume the electric field is pointing in the negative x-direction as shown in the figure.
An electron is projected with an initial speed v0 = 1.70×106 m/s into the uniform field between the parallel plates in (Figure 1). Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. Express your answer in newtons per coulomb. Suppose that in the figure the electron is replaced by a proton with the same initial speed v0v0. Would the proton hit one of the plates? What would be the direction of proton's displacement? displacement is upward displacement is downward
A positive charge q is fixed at point (3,4) and a negative charge −q is fixed at point (3,0). Determine the net electric force F→net acting on a negative test charge −Q at the origin (0,0) in terms of the given quantities and physical constants, including the permittivity of free space ε0. Express the force using ij unit vector notation. Enter precise fractions rather than entering their approximate numerical values.
a) Calculate the acceleration in meters per second squared of the electron if the field strength is 2.50 ✕ 104 N/C.
A uranium ion and an iron ion are separated by a distance of ?=57.10 nm, as shown in the figure. The uranium atom is singly ionized; the iron atom is doubly ionized. Calculate the distance ? from the uranium atom at which an electron will be in equilibrium. Ignore the gravitational attraction between the particles. ?= What is the magnitude ?U of the force on the electron from the uranium ion?
The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. 1a. Given the oil drop to be 0.99 μm in radius and have a density of 910 kg/m3: Find the weight of the drop. 1b.If the drop has a single excess electron, find the electric field strength needed to balance its weight. This is not and will not be graded
A simple and common technique for accelerating electrons is shown in the figure, which depicts a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through.Randomized VariablesE = 2.7 × 104 N/C Calculate the horizontal component of the electron's acceleration if the field strength is 2.7 × 104 N/C. Express your answer in meters per second squared, and assume the electric field is pointing in the negative x-direction as shown in the figure.
In the early 1900's Robert Millikan discovered the peculiar property that charge came in little packets, no smaller than e = 1.602 x 10-19 C -- he had measured the charge of the electron. Here's (roughly) how he did it. He removed an electron from an initially neutral droplet of oil with diameter 0.5 ??μm. In a vacuum, he positioned the droplet between two metallic plates separated by 6 mm and fiddled with the potential (voltage) across the plates until the droplet would hover against the force of gravity. Droplets of this size with +e charge would hover, but only for a particular voltage (otherwise they would sink or rise). Given the parameters stated here, and the fact that the density of the oil was 831 kg/m3, what was the voltage that made the droplets hover? (give your answer with 0.1 V precision)