An electron is projected with an initial speed v0= 1.60 ×10^6 m / s into the uniform field betw

step 1 In this question we have this set up. There's a parallel plate with uniform electric field insid

An electron is projected with an initial speed v0= 1.60...

An electron is projected with an initial speed $v_{0}=$ $1.60 \times 10^{6} \mathrm{~m} / \mathrm{s}$ into the uniform field between two parallel plates (Fig. E21.27). 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. (a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. (b) Suppose that the electron in Fig. E21.27 is replaced by a proton with the same initial speed $v_{0} .$ Would the proton hit one of the plates? If not, what would be the magnitude and direction of its vertical displacement as it exits the region between the plates? (c) Compare the paths traveled by the electron and the proton, and explain the differences. (d) Discuss whether it is reasonable to ignore the effects of gravity for each particle.

Key Concepts

Charge-to-Mass Ratio

The charge-to-mass ratio, defined as q/m, is crucial in determining the acceleration of a charged particle in an electric field. A higher charge-to-mass ratio means a greater acceleration for the same field strength. This concept explains why electrons and protons, despite having charges of equal magnitude, accelerate differently due to their vastly different masses.

Projectile Motion under Uniform Acceleration

When a particle enters a uniform electric field with an initial velocity, its motion can be described as projectile motion where one component (typically the horizontal) remains constant and the other (vertical) experiences uniform acceleration. This decomposition of motion into orthogonal components allows the use of kinematic equations to calculate displacement, time of flight, and final positions.

Negligibility of Gravitational Effects

When analyzing the motion of charged particles in a strong electric field, the influence of gravity can often be considered negligible. This is because the electric forces and resulting accelerations are typically many orders of magnitude larger than those due to gravity, allowing simplifications in calculations by ignoring gravitational effects.

Uniform Electric Field

A uniform electric field is one in which the electric force experienced by a charged particle is constant in magnitude and direction throughout the region. This concept is fundamental in understanding particle trajectories in devices like cathode ray tubes and particle accelerators, where the field creates a predictable, constant acceleration, allowing the application of kinematics to analyze motion.

Electric Force on Charged Particles

The electric force acting on a charged particle in an electric field is given by F = qE, where q is the charge and E is the electric field strength. This force results in an acceleration a = qE/m, dependent on both the charge and the mass of the particle. This concept explains how particles with different charge-to-mass ratios, such as electrons and protons, respond differently when subjected to the same electric field.

Transcript

00:01 In this question we have this set up.

00:02 There's a parallel plate with uniform electric field inside the plate, in between the plates, and then there's an electron projected in the middle of the parallel plates, okay? and then it barely escapes or misses the upper plate, okay, as it emerges from the field.

00:25 So we want to find that many two of electric fuel, okay? and then there are three more parts related to change to proton and then looking at the motion of electron and proton in this uniform electric field.

00:44 Okay, that is perpendicular to the motion.

00:48 Right, so part a, we want to find the electric field, the magnitude of the electric field.

00:56 So first we recognize that the electron is doing projectile motion.

01:03 Okay, so the path is a parabola.

01:06 Okay, and then so the horizontal motion and the vertical motion, they are not affecting each other.

01:15 Okay, okay, first we use f equals to qe equals to m .a.

01:24 Right? so the acceleration would just be e divided by m.

01:33 So this is an expression for acceleration.

01:36 And then the vertical displacement is over here.

01:43 So delta y is 0 .5 cm, right? 0 .5 times 10 to the negative 2 meters.

01:53 And this is equal to half a t square.

01:58 Because there is no initial vertical velocity.

02:03 And this equation is actually using delta y equals to b0y t plus half a t square.

02:16 Okay, so this is the equation that i'm using for this part.

02:23 And then so b0 y is 0 and delta y is 0 .5 times 10 to the negative 2 meters.

02:29 And then so you are left with half a t square.

02:32 Okay, so you can actually substitute some numbers and then we also need to find out the t.

02:42 Okay, t will just be delta x over v0, okay? because the time taken to travel across the plate will be the time taken for it to move vertically up.

02:53 Okay, so this is 0 .02, divide by 1 .6 times 10 to the 6.

03:02 So it travels out to cm horizontally.

03:07 Okay, so that's why 0 .02 meters.

03:09 Alright, so we have all the ingredients.

03:12 Okay, so 0 .5 times 10 to the negative 2.

03:16 This is a delta y equals to half e, the charge times the electric field divided by m, then t squared, okay? so i want to find the electric field.

03:30 So it would be 2m.

03:33 Times 0 .5 times 10 to the negative 2 divide by e t squared you can start substituting the numbers and then divide by 1 .6 times 10 to the 19 and then the t would be 0 .02 divided by 1 .6 times 10 to the 6 square okay you do the calculation use your calculator you get 364 words per meter.

04:12 So this is the answer for part a.

04:17 Then in part b, you change the electron to proton.

04:23 And then with the proton hit the plate.

04:27 And then if not, what would be the magnitude and direction of its vertical displacement? as it exists a few.

04:35 So just redraw the situation again.

04:38 Okay, the field is pointing down...

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