A pair of oppositely charged parallel plates are separated by 5.33 mm. A potential difference of

A pair of oppositely charged parallel plates are separated...

A pair of oppositely charged parallel plates are separated by $5.33 \mathrm{mm}$. A potential difference of $600.0 \mathrm{V}$ exists between the plates. a. What is the magnitude of the electric field strength in the region that is located between the plates? b. What is the magnitude of the force on an electron that is in the region between the plates at a point that is exactly $2.90 \mathrm{mm}$ from the positive plate? c. The electron is moved to the negative plate from an initial position $2.90 \mathrm{mm}$ from the positive plate. What is the change in electrical potential energy due to the movement of this electron?

A pair of oppositely charged parallel plates are separated by $5.33 \mathrm{mm}$. A potential difference of $600.0 \mathrm{V}$ exists between the plates.

a. What is the magnitude of the electric field strength in the region that is located between the plates?
b. What is the magnitude of the force on an electron that is in the region between the plates at a point that is exactly $2.90 \mathrm{mm}$ from the positive plate?
c. The electron is moved to the negative plate from an initial position $2.90 \mathrm{mm}$ from the positive plate. What is the change in electrical potential energy due to the movement of this electron?

Key Concepts

Uniform Electric Field

In a scenario like a parallel plate arrangement, the electric field is uniform between the plates. Its magnitude is determined by dividing the potential difference by the separation between the plates, and it defines the force experienced at every point in the region between the plates.

Electric Force on a Charge

A charged particle placed in an electric field experiences a force that is directly proportional to both its charge and the electric field strength. Mathematically, this is expressed as F = qE, where q is the charge and E is the electric field.

Electric Potential Energy

The electric potential energy of a charge in an electric field depends on its position relative to a reference point. The change in this energy when a charge moves between two positions in the field is given by the product of the charge and the change in electric potential between those points.

Potential Difference

Potential difference represents the difference in electric potential between two points. In the context of a uniform electric field, it relates directly to the magnitude of the electric field and the distance over which that field is applied.

Transcript

00:01 So if we're looking at a parallel plate capacitor here, we can think about a few things.

00:08 So one of them is the magnitude of the electric field.

00:12 We can also think about the magnitude of the force of an electron, as well as the change in potential energy.

00:19 But if i was to take an electron, say from like here and then move it to there, or vice versa.

00:27 So if we take, let's look at the first part here, so the magnitude of the electric field.

00:37 So the equation for this here is that the electric field is the change in voltage divided by the distance.

00:45 And both of these values here are given in the problem.

00:48 This is the distance between the two plates and the voltage applied between the two plates is 600 volts.

00:55 An important thing to remember about capacitors is that the electric field is the same in between.

01:01 So if i just plug in these numbers, the electric field is going to be at that strength everywhere within this parallel plate capacity.

01:10 So plugging in these numbers, i've got 600 volts divided by 5 .33 millimeters.

01:18 I need to make sure remember to put this in as meters.

01:21 So this is 5 .33 times 10 to the minus 3 meters.

01:25 And here i can just plug this into my calculator and i get out a relatively large number here.

01:33 So, yeah, 112 ,520 newtons per coulum.

01:44 Sorry, 570 newtons per coulum.

01:48 So now another thing to look at here is the magnitude of the force.

01:52 So we're going to say that the electron here is at 2 .9 millimeters from the positive plate.

02:13 So this distance here is 2 .9 millimeters, and i did not draw this to scale, evidently, where i drew the electron in there.

02:27 All right.

02:29 So in order to figure that out, i can go back to a different electric field equation here.

02:35 So e equals the force divided by the charge.

02:41 Or i can rearrange this and get force equals the charge times the electric field...

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