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[Solved] Velocity of a Proton in a Capacitor

SOLVED

1. Homework Statement

A proton is fired horizontally with a speed of 1.1 106 m/s through the parallel-plate capacitor shown in Figure P34.37. The capacitor's electric field is = (1.3 105 V/m, down) and the distance between the plates is 31 millimeter.
(Image Attached)


The proton entered the capacitor gap 18.6 milli-meters above the lower plate. Where doe the proton strike the lower plate?

Measured in meters from the front edge of the lower plate.

2. Homework Equations

Kinematics Equations

3. The Attempt at a Solution

There were other questions on this problem that I already solved for. So here are other things that I know:

The magnitude of the acceleration= 1.245E13 m/s/s

Voltage difference= 4030 V


Now I've set up a kinematics table (seperate for horizontal and verticle):

Known for horizontal: Initial V= 1.1E6; A= 1.245E13
Uknown: distance, final velocity, time

Known for verticle: x= .0186 meters; Initial v= 0
Uknown: final velocity, acceleration, time


Any suggestions? I've no idea where to go from here.

Thanks in advance!

~Phoenix
 

Attachments

Last edited:

alphysicist

Homework Helper
2,238
1
Re: Velocity of a Proton in a Capacitor

Hi Phoenixtears,

1. Homework Statement

A proton is fired horizontally with a speed of 1.1 106 m/s through the parallel-plate capacitor shown in Figure P34.37. The capacitor's electric field is = (1.3 105 V/m, down) and the distance between the plates is 31 millimeter.
(Image Attached)


The proton entered the capacitor gap 18.6 milli-meters above the lower plate. Where doe the proton strike the lower plate?

Measured in meters from the front edge of the lower plate.

2. Homework Equations

Kinematics Equations

3. The Attempt at a Solution

There were other questions on this problem that I already solved for. So here are other things that I know:

The magnitude of the acceleration= 1.245E13 m/s/s
Since the electric field is downwards, what is the direction of this acceleration? I think the answer for this will change your kinematics table below.


Voltage difference= 4030 V


Now I've set up a kinematics table (seperate for horizontal and verticle):

Known for horizontal: Initial V= 1.1E6; A= 1.245E13
Uknown: distance, final velocity, time

Known for verticle: x= .0186 meters; Initial v= 0
Uknown: final velocity, acceleration, time


Any suggestions? I've no idea where to go from here.

Thanks in advance!

~Phoenix
 
Re: Velocity of a Proton in a Capacitor

Aha! I see. That makes so much more sense.

Becasue the field is down, the acceleration is also down. Then the horizonatal acceleration is 0 (I should have listened to my sister :-p). Thank you so much!
 

alphysicist

Homework Helper
2,238
1
Re: Velocity of a Proton in a Capacitor

Aha! I see. That makes so much more sense.

Becasue the field is down, the acceleration is also down. Then the horizonatal acceleration is 0 (I should have listened to my sister :-p). Thank you so much!
Sure, glad to help!
 

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