(a) An electron is moving east in a uniform electric field of 1.50 N / C directed to the west.

(a) An electron is moving east in a uniform electric field...

(a) An electron is moving east in a uniform electric field of $1.50 \mathrm{~N} / \mathrm{C}$ directed to the west. At point $A,$ the velocity of the electron is $4.50 \times 10^{5} \mathrm{~m} / \mathrm{s}$ toward the east. What is the speed of the electron when it reaches point $B, 0.375 \mathrm{~m}$ east of point $A ?$ (b) A proton is moving in the uniform electric field of part (a). At point $A,$ the velocity of the proton is $1.90 \times 10^{4} \mathrm{~m} / \mathrm{s},$ east. What is the speed of the proton at point $B ?$

Key Concepts

Work-Energy Theorem

The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. In the context of charged particles in electric fields, the work done by the electric force as the particle moves a certain distance is converted into changes in the particle's kinetic energy, allowing one to compute changes in speed as the particle moves within the field.

Electric Force on Charged Particles

This concept explains that a charged particle in an electric field experiences a force given by F = qE, where q is the charge of the particle and E is the electric field. The direction of the force depends on the sign of the charge: positive charges are accelerated in the direction of the field, while negative charges are accelerated opposite to the field direction.

Uniform Electric Field

A uniform electric field is one in which the magnitude and direction of the field remain constant throughout the region of interest. In such fields, the force acting on a charged particle is constant, resulting in uniformly accelerated motion, which simplifies the analysis of the particle’s kinematics using energy methods or kinematic equations.

Transcript

00:01 The first step to doing this problem is to figure out what the force is due to the electric field.

00:07 And so we're going to use the equation f is equal to absolute value of the charge times the electric field.

00:15 In the case of an electron, the force will be opposite to the electric field, since the charge will be negative.

00:22 And so it'll actually speed up the electron along this path here that we're focused on.

00:27 Plugging in q gives 1 .602 times 10 to the minus 19 coulums.

00:35 And then multiplied by the electric field they give us, which is 1 .5 nons per cool.

00:41 I get a force of 2 .403 times 10 to negative 19 newtons.

00:49 And again, this is positive because if the positive x direction is this way, this force will speed up the electron.

00:58 After we find the force, we can calculate the acceleration by taking the force and dividing by the mass.

01:05 In this case, we're dividing by the mass of the electron.

01:09 And so plugging this and this in gives us an acceleration of positive 2 .638 times 10 to the 11 meters per second squared.

01:20 So an extremely fast acceleration.

01:24 Now that we know the acceleration, we can do some kinematic equation to figure out what the final velocity is.

01:31 So if the initial velocity is positive 4 .5 times 10 to the 5th, i'm just going to avoid using units because they're all nsi units already.

01:40 The acceleration is here.

01:42 We know the distance travel, x minus x not, is equal to 0 .375.

01:49 And we want to figure out what the final velocity is.

01:52 To do this, we're going to use the kinemak equation, v squared, is equal to v0 squared, plus 2ax minus x not.

02:01 Solving this for v gives 6 .33 times 10 to the fifth and now i'll add in the units here...

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