# An electron is fired at 4.0x10^6 m/s horizontally....

• LionLieOn
In summary, an electron is fired at 4.0x10^6 m/s between parallel plates with an electric field of 4.0x10^2 N/C. The separation between the plates is 2.0 cm. The acceleration of the electron is 0 m/s^2 and the horizontal distance traveled is 0 m. To find the velocity of the electron as it strikes the bottom plate, the distance between the plates (0.02 m) should be used instead of the distance traveled by the electron. The final velocity can be found using the equation V^2 = V2(y)^2 + V1(x)^2 and should be specified with both magnitude and direction or in unit vector notation.
LionLieOn

## Homework Statement

An electron is fired at 4.0x10^6 m/s horizontally between the parallel plates as shown, (see attachment) starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x10^2 N/C. The separation of the plates is 2.0 cm

A) find the acceleration of the electron
b) find the horizontal distance traveled by the electron when it hits the plate
c) find the velocity of the electron as it strikes the plate

## The Attempt at a Solution

Please see attachments for answers A and B.

I'm having trouble with question C.

I'm using the equation, " V2^2 = V1^2 + 2a (d) " to find V2 however, I'm not too sure which "d" I should use, the distance between the 2 plates (2.0cm) or the distance I found for question B. Can someone give me an explanation as to which one I need to use and why?

Or am I using the wrong formula/ Approaching the question wrong ?

#### Attachments

• HWQ.jpg
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• HWELECA.jpg
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• HWELECB.jpg
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Which component of the velocity are you trying to calculate with your equation? Horizontal or vertical?

Assuming x-axis to be parallel to the plates, there is no force in the x direction, therefore acceleration is zero. Apply equation of motion to find vx, distance here will be dx. There is acceleration in y direction, so calculate vy accordingly. Calculate net velocity.

PeroK said:
Which component of the velocity are you trying to calculate with your equation? Horizontal or vertical?
Vertical.

I'm assuming I use the distance between the plates (2.0 cm = 0.02 m) since I'm trying to find the Vertical component. It wouldn't make any sense if I use 0.096m since that's horizontal.

Or am I wrong?

LionLieOn said:
Vertical.

I'm assuming I use the distance between the plates (2.0 cm = 0.02 m) since I'm trying to find the Vertical component.

You should be able to do better than assume. Think what would happen if the horizontal velocity and hence displacement was 0.

PeroK said:
You should be able to do better than assume. Think what would happen if the horizontal velocity and hence displacement was 0.

Hmm I'll try my best not to assume in regards to your question.

If horizontal velocity and hence displacement were 0,
- Vertical acceleration would remain the same 7.03 X 10^13
- Time would also remain the same
- Vertical Velocity would also remain the same since

V2y^2 = V1y^2 + 2ay (dy)

V1y would still be 0.

So to answer my own question, the Distance I would use is the distance between the plates not the distance the electron has traveled. If i were to use the distance the electron has traveled, it would go against the equation making it,
V2y^2 = V1y^2 + 2ay (d "x" ) which wouldn't make sense.

Now, to finish off Question C I would find V2y and then use

V^2 = V2(y)^2 + V1(x)^2

to find the velocity of the electron as it strikes the plate, and of course, look for the direction as well.

Is this correct?

Last edited:
LionLieOn said:
So to answer my own question, the Distance I would use is the distance between the plates not the distance the electron has traveled. If i were to use the distance the electron has traveled, it would go against the equation making it,
V2y^2 = V1y^2 + 2ay (d "x" ) which wouldn't make sense.

Now, to finish off Question C I would find V2y and then use

V^2 = V2(y)^2 + V1(x)^2

to find the velocity of the electron as it strikes the plate.

Is this correct?
Yes. But velocity is a vector quantity. So, you need to either specify the answer with both a magnitude and a direction, or give the answer in unit vector notation.

TSny said:
Yes. But velocity is a vector quantity. So, you need to either specify the answer with both a magnitude and a direction, or give the answer in unit vector notation.

Agreed. I didn't include direction in my last post just because I thought it was self explanatory. I shall edit it to make it more correct.

## 1. What is the initial velocity of the electron?

The initial velocity of the electron is 4.0x10^6 m/s horizontally.

## 2. What does the horizontal direction refer to in this scenario?

The horizontal direction refers to the direction in which the electron is moving side to side.

## 3. How is the speed of the electron measured?

The speed of the electron is measured in meters per second (m/s).

## 4. What factors can affect the speed of the electron?

The speed of the electron can be affected by external forces such as electromagnetic fields and collisions with other particles.

## 5. What is the significance of the notation 4.0x10^6 m/s?

The notation 4.0x10^6 m/s represents scientific notation, meaning the electron is traveling at a speed of 4.0 multiplied by 10 to the power of 6 meters per second. This notation is used to represent large numbers in a more concise and standardized way.

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