consider a uniform electric field of magnitude e0 pointing in the j direction as we will see lat

Step 1: Start with the given information that the electric field is pointing in the -y direction

Consider a uniform electric field of magnitude e0 pointing...

Consider a uniform electric field of magnitude E0 pointing in the -ŷ direction. As we will see later in this course, one way to create such a field is to use two very large and oppositely charged plates. The plates are contained in the (x,z) plane, are of length L in the ŷ direction, and are separated by a distance 2d (the dimension in the ŷ direction is irrelevant in this problem). A point particle of mass m and positive charge q is initially outside the gap between the two plates moving along the +x axis with a speed of v0. The particle enters the region of a uniform electric field at the midpoint between the two plates. Calculate the minimum speed v0 needed for the particle not to hit the lower plate. Express your answer in terms of E0, L, d, q, and m.

Transcript

00:01 Okay, so we've got two parallel plates.

00:03 It says they're in the xz plane.

00:06 So let's just draw them like this, where this i'm saying is the x direction.

00:14 Z is into the page and y is up.

00:17 They've got negligible thickness in the y direction.

00:20 But it says they're separated by a distance 2d and they have length l in the x direction.

00:29 They say there's an electric field pointing in the negative.

00:34 Now, i can't see, it hasn't transcribed well, so it says in the minus question mark direction.

00:39 So i'm just going to take that to be down like this.

00:44 So i'm going to say this is where the electric field's pointing, i .e.

00:46 The top plate is positively charged and the bottom plate is negatively charged.

00:51 Now we have a particle that comes in in the middle, a positively charged particle, and we want to find the initial velocity it needs to have.

01:01 Is travelling wholly in the x direction, the question tells us when it comes in, such that it just escapes the plate, so the minimum speed to just escape the lower plates.

01:13 So we're going to use that here.

01:14 So we're going to use our displacement equations.

01:17 So there's no acceleration in the x direction because the electric field is pointing solely downwards, which means that integrating this to find the velocity in the x direction, we get that the velocity in the x direction is always just the initial velocity.

01:32 Which it's called v0.

01:36 And therefore the displacement in the x direction is v0t plus what we'll call sx, not the initial displacement.

01:45 But i'm just going to define this point just before it enters the plate to be the origin.

01:53 So we're going to set this to be zero.

01:55 So the horizontal displacement is just v0 .t.

02:00 And we can use this to find that the time when it it passes this point here, when it passes the end of the plate, is just going to be l, which is the horizontal displacement of the plates divided by v -0.

02:18 We want to do the same thing for the vertical component now.

02:20 So the vertical component is going to accelerate this positive charge downwards with a force, according to newton's law, of minus fy over m...

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