A proton is traveling horizontally to the right at 4.50 ×10^6 m/s. (a) Find the magnitude and di

step 1 Hey everyone. This is question number 25 from chapter 21. In this problem, we're given a proton

A proton is traveling horizontally to the right at 4.50...

A proton is traveling horizontally to the right at 4.50 $\times 10^6$ m$/$s. (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 cm. (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Key Concepts

Uniform Kinematics

This concept involves the use of kinematic equations to describe the motion of particles experiencing constant acceleration. It allows the determination of displacement, time, and velocity when acceleration is uniform, which is fundamental in analyzing how the proton slows down over a given distance.

Electrostatic Force

Charged particles in an electric field experience a force proportional to the magnitude of the field and the charge of the particle, as given by F = qE. This force is responsible for accelerating or decelerating the particle depending on the direction of the electric field relative to the charge.

Newton’s Second Law

Newton’s Second Law (F = ma) connects the net force acting on an object with its acceleration. In the context of charged particle motion in an electric field, it is used to relate the electrostatic force exerted on the particle to its mass and the resulting acceleration or deceleration.

Charge Sign and Field Direction

The sign of the charge determines the direction of the force experienced in a given electric field. Positive charges, like the proton, and negative charges, such as the electron, react differently to the same field direction, which needs to be taken into account when calculating the field necessary to bring them to rest.

Transcript

00:01 Hey everyone.

00:01 This is question number 25 from chapter 21.

00:04 In this problem, we're given a proton that is traveling horizontally to the right at 4 .5 times 10 to the 6 meters per second.

00:11 We're asked to find the magnitude and direction of the weakest electric field that can bring the proton to rest.

00:17 We're asked to find the time it takes for the proton to stop.

00:20 And then we're asked to find the minimum field magnitude and direction needed to stop an electron.

00:25 So i sketch us out.

00:26 This is the proton.

00:27 It's moving to the right.

00:28 So for part a here, magnitude and direction, of the weakest electron electric field that can bring the proton to rest.

00:36 If we're looking at a proton moving to the right, it's going to have to stop, it's going to have to decelerate and stop moving, and that means the electric field is going to have to point to the left.

00:46 So we can go ahead and establish that before we even do any math.

00:50 But let's move on to the math here.

00:54 We have a distance and we have a speed.

00:57 So we have distance, we have velocity.

00:59 We have a final velocity of zero, but in order to find an electric field, we need some more information because the equations that we have for electric fields are f equals eq, and we know that f equals m .a.

01:26 So we know mass of a proton, but we don't have a here.

01:31 And we know charge, and we need to find this electric field, so what we need to find first versus a.

01:35 And we can actually do that with kinematics, throwing it back a little bit.

01:40 So we have a kinematic equation that relates velocity and distance, and that is v squared equals v0 squared plus 2 ad.

01:55 So we have our v .0, our vf obviously, is zero, because we're coming to a stop and we have our distance.

02:01 So we can rearrange this equation and solve for a.

02:09 And we get a equals, we subtract v0 over to the other side.

02:14 Side so minus v0 squared over 2d which is our distance and we're given that in the problem so now we can go ahead and plug in and solve for everything so our v knot is minus 4 .5 that's squared and we divide that by two and our distance here which is um let's see 33 .2 centimeters so we have to move that over and we get 0 .032 meters and that's going to give us an acceleration of minus 3 .164 times 10 to the 14 meters per second squared.

03:31 So that's our acceleration.

03:35 And now we look back up at these equations that i wrote f equals eq...

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