Vertical Deflection of Electron: 1.1×107 m/s & 3.2×10-16 N

[J-dizzal]

Homework Statement

An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.2 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 34 mm horizontally.

Homework Equations

F=ma[/B]

The Attempt at a Solution

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[Matternot]

Try starting by using your "relevant equation"?

When calculating the distance travelled, what information do we need to calculate initially?

 

[J-dizzal]

Try starting by using your "relevant equation"?

When calculating the distance travelled, what information do we need to calculate initially?

distance = vt?

 

[Matternot]

That's not a bad idea. We will need that equation. The problem is the velocity isn't constant because there is a force. How do we calculate the time it is traveling for though? The time it's traveling for is important.

 

[J-dizzal]

v = v₀ + at and solve for t? this eqate to zero, nevermind

 

[Matternot]

It's very important to treat the vertical and horizontal velocities separately. The vertical force does not affect the horizontal speed. The horizontal speed does not affect the vertical speed. Maybe try using $s=vt$?

 

[J-dizzal]

That's not a bad idea. We will need that equation. The problem is the velocity isn't constant because there is a force. How do we calculate the time it is traveling for though? The time it's traveling for is important.

would i use a motion equation(constant accerleration)?

 

[Matternot]

Yeah, The motion equation for constant acceleration sounds like a really good idea to describe the vertical motion. The problem is we need to calculate the time it is traveling for to use that. The horizontal motion might help us here.

 

[J-dizzal]

It's very important to treat the vertical and horizontal velocities separately. The vertical force does not affect the horizontal speed. The horizontal speed does not affect the vertical speed. Maybe try using $s=vt$?

I don't see how a vertical force would'nt affect the horizontal speed. for example if the vertical force was enough to change the direction of the particle to a vertical path then its horizontal speed would be zero. and the horizontal speed would be changing all the way.

 

[Matternot]

But what if the force can never be strong enough to change the direction of the particle to a vertical path? What if the horizontal speed can never be zero under the influence of this vertical force?

 

[J-dizzal]

But what if the force can never be strong enough to change the direction of the particle to a vertical path? What if the horizontal speed can never be zero under the influence of this vertical force?

well then the vertical force is not strong enough, but even a very small vertical force acting on a horizontal particle would accelerate the particle vertically therefore changing its horizontal velocity. the speed of particle would not change, just the horizontal velocity dependent on its angle with respect to horizontal.

 

[J-dizzal]

But what if the force can never be strong enough to change the direction of the particle to a vertical path? What if the horizontal speed can never be zero under the influence of this vertical force?

so, distance = vt and solving for t would not work yet because the time isn't known. and distance is not known

 

[Matternot]

Why is the speed of the particle constant?

 

[J-dizzal]

Why is the speed of the particle constant?

speed is constant assuming this is in a vacuum, sorry should of noted that

 

[Matternot]

so, distance = vt and solving for t would not work yet because the time isn't known. and distance is not known

This will make sense when the separation of x and y directions makes sense. I think we should postpone this until that is understood.

I assume you're trying to use the idea of constant energy. Why do you think energy here might not be constant?

 

[J-dizzal]

i do not know about energy, i don't thing we have got to that chapter in the book yet.

 

[Matternot]

So what makes you think the speed is constant?

 

[J-dizzal]

So what makes you think the speed is constant?

so would the particle be accelerating after it enters the force field?

 

[J-dizzal]

So what makes you think the speed is constant?

wouldnt speed be constant because there are no other forces acting on the particle that are past 90deg

 

[Matternot]

Where there is a net force on an object, there is always an acceleration. Newton's first.

The idea that horizontal and vertical motion can be treated separately is crucial.

Imagine throwing a ball to someone. When it is thrown it is given a horizontal velocity and a vertical velocity. The gravitational force will affect it's vertical velocity, but not the horizontal. As it's vertical velocity is being affected, the direction and speed of the ball is being affected.

 

[Matternot]

Instantaneously no work is being done as the force is perpendicular to the ball. The speed initially is constant. The direction of the ball however slowly changes. Therefore the angle the force makes with the velocity of the ball changes (i.e. not to 90°). It therefore begins to speed up.

 

[J-dizzal]

Where there is a net force on an object, there is always an acceleration. Newton's first.

The idea that horizontal and vertical motion can be treated separately is crucial.

Imagine throwing a ball to someone. When it is thrown it is given a horizontal velocity and a vertical velocity. The gravitational force will affect it's vertical velocity, but not the horizontal. As it's vertical velocity is being affected, the direction and speed of the ball is being affected.

Ok, so I am missing a force which is gravity? also air resistance is omitted in my problem

 

[Matternot]

Gravity will be negligible in this question as mg<<F. There is no need to include it. The problem given however is analogous to a ball being thrown.

Instead consider the position of a particle in space.

The x co-ordinate has no effect on the y co-ordinate.
Therefore the x velocity has no effect on the y co-ordinate.
Therefore the x acceleration has no effect on the y co-ordinate.

Do you see where I'm going with this?

 

[J-dizzal]

Instantaneously no work is being done as the force is perpendicular to the ball. The speed initially is constant. The direction of the ball however slowly changes. Therefore the angle the force makes with the velocity of the ball changes (i.e. not to 90°). It therefore begins to speed up.

Ok then the electron has an initial velocity and its also accelerating due to the vertical force field. so its velocity in the y direction is changing. So then i have to find acceleration of the electron due to the force which is a = f/m = 2.5125x10¹⁴ and this is the for the y component?

 

[J-dizzal]

Ok then the electron has an initial velocity and its also accelerating due to the vertical force field. so its velocity in the y direction is changing. So then i have to find acceleration of the electron due to the force which is a = f/m = 2.5125x10¹⁴ and this is the for the y component?

this is the acceleration of the electron in the y dirction (vertical).

 

[Matternot]

I think you might have a type in the calculation. I calculate a slightly different acceleration.

precisely though. This is for the y component of the velocity.

Lets think instead about the horizontal velocity. What is the acceleration (if any) in the x direction?

 

[J-dizzal]

I think you might have a type in the calculation. I calculate a slightly different acceleration.

precisely though. This is for the y component of the velocity.

Lets think instead about the horizontal velocity. What is the acceleration (if any) in the x direction?

there must be acceleration in the -x direction

 

[Matternot]

Why?

 

[J-dizzal]

Why?

because the electron is changing direction, the x component of its velocity is getting smaller and the y component is getting larger?

 

[J-dizzal]

Why?

maybe not, if there are no forces acting in the -x direction, then the x component of its velocity doesn't change. its only the y component increasing right?