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

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

An electron is projected with an initial speed $v_0 =$ 1.60 $\times$ $10^6$ m$/$s into the uniform field between two parallel plates ($\textbf{Fig. E21.29}$). 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. E21.29 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

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.

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.

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.

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.

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.

Transcript

00:01 So while traveling through the electric field, the electron follows a projectile path like so, with x0 and x ,0 and y0 as the initial position of the electron, and x, f, y, f as the final position.

00:21 So as you can see from the graph here, x, not is equal to zero, y, f, y, f as is negative 0 .005 meter xf is 0 .02 meter which is the total length of this electric field and yf is 0.

00:53 Y f is 0.

00:58 Now for the first part we have to find the electric field so for that we need to find the acceleration and in order to find the acceleration, we need to find the time at which it reaches the end of the plates.

01:18 So basically we know that electric field is equal to force over charge and force by newton's law is mass times acceleration over charge.

01:36 And once we know the acceleration, we can calculate the electric field.

01:42 So let's use kinematic equations to find the xerlars.

01:46 Acceleration.

01:48 So first we need to calculate the time.

01:52 So this is the equation that we will be using.

02:00 So this is the time taken by the electron in order to travel this distance.

02:07 Now since there is no acceleration in the horizontal direction, this is equal to 0 because a x is zero and acceleration is in vertical direction is gravity, which is 9 .8 meter per second.

02:26 Negative 98 point meter per second because gravity points downward.

02:32 Also, the projectile has the fallen values for velocity.

02:37 So v0x is 1 .6 times 10 to the power 6 meter per second.

02:44 And since there is no acceleration in the x direction, the final velocity along x direction or the component of final velocity along x direction is the same as the initial one.

03:03 And v0y is equal to 0 because there is no initial component of velocity along y direction.

03:12 Now, from this equation, we have t to be equal to x minus x0 over v0x.

03:20 And substituting these values, we get time to be equal to 1 .25 times 10 to the power minus 8 seconds.

03:29 So now we can find the vertical acceleration, a y.

03:35 So remember this is actually m -a -y because the electric field is along y direction.

03:43 So the force we want to calculate should be the vertical force, and therefore we need to find the vertical acceleration.

03:51 So we will use this same kinematic equation, but solve it for the vertical direction.

03:57 So we have y equal to y -0 -0 -1.

03:59 Plus v0y t plus half a y delt squared.

04:07 So in this equation v0 y is 0, so that term goes away.

04:14 And therefore we have a y to be equal to y minus y not times 2 over delta t square.

04:26 Also the capital, the final value of y is 0.

04:31 So this also goes to zero.

04:36 So finally we have negative twice of y -not over delta t squared.

04:47 So now we can substitute the value of time that we just calculated and the value of initial y and find the acceleration.

04:57 So that comes out to be equal to 6 .4 times 10 to the bar 13.

05:05 Meter per second square.

05:07 So now that we have a y, we can plug this into this equation where we have the electric field in terms of acceleration.

05:17 So ey is equal to m -a -y over the charge.

05:29 And here we have already taken into account gravity or we are ignoring gravity.

05:36 So we will find that in the last part.

05:42 So this value of gravity, we are not considering gravity here...

Please give Ace some feedback