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

[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

Homework Equations

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 ?

 

[PeroK]

Which component of the velocity are you trying to calculate with your equation? Horizontal or vertical?

 

[cheapstrike]

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 v_x, distance here will be d_x. There is acceleration in y direction, so calculate v_y accordingly. Calculate net velocity.

 

[LionLieOn]

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?

 

[PeroK]

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.

 

[LionLieOn]

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?

 

[TSny]

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.

 

[LionLieOn]

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.