An electron has an initial velocity of v⃗0⃗ = -20,000 m/s ĵ. It is positioned near two large pla

Step 1: The electron is moving in the y-direction only, so its x-position remains constant.

An electron has an initial velocity of v⃗0⃗ = -20,000 m/s ĵ....

An electron has an initial velocity of \(\vec{v_0} = -20,000\text{ m/s }\hat{j}\). It is positioned near two large plate charges which appear as straight lines in the figure (Figure 1). The charge per unit area of each plate is given in the figure. What is the x position of the electron after 3 ns? Round your answer to three significant figures. Figure 0.5 nC/m² • electron -0.3 nC/m² What is the final y position of the electron? Round your answer to three significant figures.

Transcript

00:01 Okay, so let's say we have two plates with an electric field in between them, and it points upwards, we're told.

00:07 And the electric field has a magnitude of 400 newtons per coulom.

00:15 And we're told that an electron travels in at a very high speed and is deflected downwards and hits the plates.

00:21 So the first part of this question, part a, says, what is the sign of the plates? well, the electric field is going to go from positive to negative.

00:31 And so the bottom plate has to be positively charged.

00:34 The top plate has to be negatively charged.

00:39 So, like, the sign of the bottom plate is positive.

00:42 The sign of the top plate is going to be negative.

00:46 So that's part a.

00:47 Part b says determine the acceleration of the electron between the plates.

00:51 Well, the force of the electron that it experienced is going to be its charge times the electric field strength.

00:57 We're kind of ignoring gravity here, but we could include gravity if we wanted to, but it's going to be pretty small.

01:04 The gravitational acceleration will be very small compared to the acceleration it feels from the electric field.

01:16 And since the electron is negatively charged, we can write this as like negative e, which is the electron charge, times the electric field, and then we'll say minus mg.

01:25 So both forces are going down.

01:26 So then the acceleration is going to be all this divided by the mass.

01:30 And the magnitude of the acceleration, it's going to be like the electron charge times the electric field plus the gravitational acceleration, because they're both in the same direction.

01:40 So the electron charge is 1 .6 times 10 to the negative 19th coulams times 400 newtons per coulomb divided by our electron mass, which is very small, 9 .11 times 10 to the negative 31st kilograms.

01:54 And then we're adding to this 9 .81 meters per second squared.

02:02 But as you'll see, the first acceleration is so big that hardly makes a difference if we divide by, if we add our secondary force, because our acceleration is basically 7 .025 times 10 to the 13th meters per second squared.

02:20 So you can see adding 9 .8 doesn't really change that answer much.

02:25 Part c, determine the horizontal distance traveled by the electron when it hits the plate.

02:30 So the electron enters a distance of two centimeters...

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