Calculate the speed of the electrons as they enter the gap

[moenste]

Homework Statement

Two parallel metal sheets of length 10 cm are separated by 20 mm in a vacuum. A narrow beam of electrons enters symmetrically between them as shown.

When a PD of 1000 V is applied between the plates the electron beam just misses one of the plates as it emerges.

Calculate the speed of the electrons as they enter the gap. (Take the field between the plates to be uniform.)

(e / m = 1.8 * 10¹¹ C kg^-1.)

Answer: 6.7 * 10⁷ m s^-1.

2. The attempt at a solution
I used 0.5 m v² = e E → v = √ 2 * (e / m) * E = √ 2 * (1.8 * 10¹¹) * 1000 = 18 973 665.96 m s^-1.

But this formula does not include the length of the plates and their separation. Not sure what to do next.

 

[PeroK]

What is $v$ in your equation for energy?

 

[moenste]

What is $v$ in your equation for energy?

Speed, velocity. Isn't it?

 

[PeroK]

Speed, velocity. Isn't it?

If you mean the speed as it enters the apparatus, what does that have to do with the strength of the electric field?

 

[cnh1995]

Homework Statement

Two parallel metal sheets of length 10 cm are separated by 20 mm in a vacuum. A narrow beam of electrons enters symmetrically between them as shown.

When a PD of 1000 V is applied between the plates the electron beam just misses one of the plates as it emerges.

Calculate the speed of the electrons as they enter the gap. (Take the field between the plates to be uniform.)

(e / m = 1.8 * 10¹¹ C kg^-1.)

Answer: 6.7 * 10⁷ m s^-1.

2. The attempt at a solution
I used 0.5 m v² = e E → v = √ 2 * (e / m) * E = √ 2 * (1.8 * 10¹¹) * 1000 = 18 973 665.96 m s^-1.

But this formula does not include the length of the plates and their separation. Not sure what to do next.

The situation in this problem is like when you throw a ball horizontally (parallel to ground) from top of a building and it hits the ground.
Which force is acting on the electron? In which direction?

 

[moenste]

If you mean the speed as it enters the apparatus, what does that have to do with the strength of the electric field?

Don't we need to find the speed?

The situation in this problem is like when you throw a ball horizontally (parallel to ground) from top of a building and it hits the ground.
Which force is acting on the electron? In which direction?

F = m g vertically downwards? And also the acceleration force which is horizontal F = m a?

 

[PeroK]

Don't we need to find the speed?F = m g vertically downwards? And also the acceleration force which is horizontal F = m a?

Post #5 above has given you a big clue. Do you understand what he is saying?

 

[moenste]

Post #5 above has given you a big clue. Do you understand what he is saying?

Is this correct:

F = m g vertically downwards? And also the acceleration force which is horizontal F = m a?

?

 

[PeroK]

Is this correct:

?

It doesn't say the plates are horizontal. In any case, the electron is moving so fast that gravity is negligible over such a short time.

Or, perhaps more accurately, once you calculate the electric force, you will see that the gravitational force is negligible by comparison.

 

[lychette]

Do you know how to calculate acceleration of an electron in an electric field?

 

[cnh1995]

@moenste, As peroK has pointed out, gravitational pull will be negligible here. The electron is in an electric field. What is the force on the electron? In which direction? Which component of its velocity will be affected by this force? We have discussed a similar problem in another thread recently.

 

[moenste]

Do you know how to calculate acceleration of an electron in an electric field?

a = E e / m = (1.8 * 10¹¹) * 1000 = 1.8 * 10¹⁴ m s^-2.

@moenste, As peroK has pointed out, gravitational pull will be negligible here. The electron is in an electric field. What is the force on the electron? In which direction? Which component of its velocity will be affected by this force? We have discussed a similar problem in another thread recently.

F = e E in an electric field. On the picture they are moving downwards, so downwards is directed the force?

 

[cnh1995]

a = E e / m = (1.8 * 10¹¹) * 1000 = 1.8 * 10¹⁴ m s^-2.F = e E in an electric field. On the picture they are moving downwards, so downwards is directed the force?

Yes. The force is acting vertically downwards.
You know the acceleration of the electron. So which component of its velocity gets affected by this force? What can you say about the displacement of the electron?

 

[moenste]

So which component of its velocity gets affected by this force? What can you say about the displacement of the electron?

Vertical component has that acceleration.

Well, it moves from left to right and is moving closer to the bottom.

 

[cnh1995]

Vertical component has that acceleration.

Well, it moves from left to right and is moving closer to the bottom.

Right. So there is a force in the vertical direction and there is no force in the horizontal direction. What does this tell you about the horizontal component of its velocity? Also, what are the displacements of the electron in vertical and horizontal directions?

 

[lychette]

a = E e / m = (1.8 * 10¹¹) * 1000 = 1.8 * 10¹⁴ m s^-2.F = e E in an electric field. On the picture they are moving downwards, so downwards is directed the force?

With this acceleration can you calculate the time taken to travel 10mm

 

[moenste]

With this acceleration can you calculate the time taken to travel 10mm

s = v t + 0.5 a t²
0.1 = 0 * t + 0.5 * (1.8 * 10¹⁴) * t²
t = 3.33 * 10^-8 s.

Right. So there is a force in the vertical direction and there is no force in the horizontal direction. What does this tell you about the horizontal component of its velocity? Also, what are the displacements of the electron in vertical and horizontal directions?

There is no horizontal component of the velocity?

Like y = 0.5 * (e E / m) * t², x = v t so y = (e E / 2 m v²) x²?

 

[lychette]

s = v t + 0.5 a t²
0.1 = 0 * t + 0.5 * (1.8 * 10¹⁴) * t²
t = 3.33 * 10^-8 s.There is no horizontal component of the velocity?

Like y = 0.5 * (e E / m) * t², x = v t so y = (e E / 2 m v²) x²?

the only force is due to the electric field which is at right angles to the 'horizontal' velocity.
You now know the time for the electron to 'fall' 10mm.
can you calculate the velocity of the electron to travel the length of the plates? (watch the units!)

 

[cnh1995]

There is no horizontal component of the velocity?

There is. How do you think the electron undergoes horizontal displacement then? No force in the horizontal direction means horizontal velocity is constant throughout (Newton's first law).

x = v t

Right. You know the horizontal displacement and time. You can calculate the horizontal component of velocity using this equation.
This is the velocity of the electron while entering the gap.

 

[gneill]

a = E e / m = (1.8 * 10¹¹) * 1000 = 1.8 * 10¹⁴ m s^-2.

E is the electric field, not the voltage. The potential difference between the plates is not the same thing as the electric field strength. You need to determine E first.

 

[moenste]

There is. How do you think the electron undergoes horizontal displacement then? No force in the horizontal direction means horizontal velocity is constant throughout (Newton's first law).

Right. You know the horizontal displacement and time. You can calculate the horizontal component of velocity using this equation.
This is the velocity of the electron while entering the gap.

x = 3 000 000 * 3.33 * 10^-8 = 0.0999 m s^-1?

I actually think that my time calculation is wrong. s = v t + 0.5 a t² and v = 0 so s = 0.5 a t² -- is it correct?

E is the electric field, not the voltage. The potential difference between the plates is not the same thing as the electric field strength. You need to determine E first.

E = V / d = 1000 / 0.02 = 50 000 V?

 

[cnh1995]

s = v t + 0.5 a t2 and v = 0 so s = 0.5 a t2 -- is it correct?

Yes.

x = 3 000 000 * 3.

x is the horizontal displacement, not velocity. Use v=x/t.

 

[gneill]

E = V / d = 1000 / 0.02 = 50 000 V?

Your units are volts divided by meters: V/m. (Or N/C, which amounts to the same thing)

Potential difference is specified in volts. The electric field is specified in volts per meter, or V/m.

 

[lychette]

I think your calculation of the electric field is out by a factor of 2...it looks like you took the separation of the plates to be 10mm !

My correction, I read the voltage as 2000V...disregard the above !

 

[moenste]

E is the electric field, not the voltage. The potential difference between the plates is not the same thing as the electric field strength. You need to determine E first.

I think your calculation of the electric field is out by a factor of 2...it looks like you took the separation of the plates to be 10mm !

My correction, I read the voltage as 2000V...disregard the above !

Hm, the length of the plates is 10 cm or 0.1 m and their separation is 20 mm or 0.02 m. What 10 mm do you mean?

Your units are volts divided by meters: V/m. (Or N/C, which amounts to the same thing)

Potential difference is specified in volts. The electric field is specified in volts per meter, or V/m.

So: E = V (voltage) / d (separation of the plates) = 1000 / 0.02 = 50 000 V m^-1. I think this should be correct.

Yes.

x is the horizontal displacement, not velocity. Use v=x/t.

1. E = V / d = 1000 / 0.02 = 50 000 V m^-1.
2. a = E * (e / m) = 50 000 * 1.8 * 10¹¹ = 9 * 10¹⁵ m s^-2.
3. s = v t + 0.5 a t², v = 0, s = 0.1 m → t = √ s / 0.5 a = √ 0.1 / 0.5 * 9 * 10¹⁵ = 4.7 * 10^-9 s.
4. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (4.7 * 10^-9)² = 0.1 m s^-1.

What's next? We can't find the x component of velocity because in [x = v t] velocity is unknown. We can't plug everything into [y = (e E / 2 m v²) x²] since we don't know x and v.

 

[gneill]

So: E = V (voltage) / d (separation of the plates) = 1000 / 0.02 = 50 000 V m^-1. I think this should be correct.

Yes, much better.

 

[gneill]

You have the vertical acceleration and the distance traveled vertically while the electron is between the plates. How long does it take for the electron to travel that distance?

 

[moenste]

You have the vertical acceleration and the distance traveled vertically while the electron is between the plates. How long does it take for the electron to travel that distance?

1. E = V / d = 1000 / 0.02 = 50 000 V m^-1.
2. a = E * (e / m) = 50 000 * 1.8 * 10¹¹ = 9 * 10¹⁵ m s^-2.

3. s = v t + 0.5 a t², v = 0, s = 0.1 m → t = √ s / 0.5 a = √ 0.1 / 0.5 * 9 * 10¹⁵ = 4.7 * 10^-9 s.
4. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (4.7 * 10^-9)² = 0.1 m s^-1.

5. x = v t → v = x / t = 0.1 / 4.7 * 10^-9 = 21 276 596 m s^-1.

6. t = √ s / 0.5 a = √ 0.02 / 0.5 * 9 * 10¹⁵ = 2.1 * 10^-9 s.
7. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (2.1 * 10^-9)² = 0.02 m s^-1.

8. v = x / t = 0.02 / 2.1 * 10^-9 = 9 523 809 m s^-1.

 

[PeroK]

1. E = V / d = 1000 / 0.02 = 50 000 V m^-1.
2. a = E * (e / m) = 50 000 * 1.8 * 10¹¹ = 9 * 10¹⁵ m s^-2.

3. s = v t + 0.5 a t², v = 0, s = 0.1 m → t = √ s / 0.5 a = √ 0.1 / 0.5 * 9 * 10¹⁵ = 4.7 * 10^-9 s.
4. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (4.7 * 10^-9)² = 0.1 m s^-1.

5. x = v t → v = x / t = 0.1 / 4.7 * 10^-9 = 21 276 596 m s^-1.

6. t = √ s / 0.5 a = √ 0.02 / 0.5 * 9 * 10¹⁵ = 2.1 * 10^-9 s.
7. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (2.1 * 10^-9)² = 0.02 m s^-1.

8. v = x / t = 0.02 / 2.1 * 10^-9 = 9 523 809 m s^-1.

This might be a radical suggestion, but what if you did this algebraically, rather than plugging in all those numbers at the first opportunity? Maybe it's easier to keep track of a few symbols than all those numbers and maybe this means it's easier to manage the units?

You have the voltage $V$, the distance between the plates $d$, the length of the plates $D = 5d$ and the initial horizontal speed of the electron $u$. And, of course, the charge $q$ and mass $m$ of the electron.

Perhaps you could shoot for the solution:

$u = 5 \sqrt{Vq/m}$

Then, as you know $V$ and $q/m$, the numerical calculation is simple.

 

[gneill]

1. E = V / d = 1000 / 0.02 = 50 000 V m^-1.
2. a = E * (e / m) = 50 000 * 1.8 * 10¹¹ = 9 * 10¹⁵ m s^-2.

3. s = v t + 0.5 a t², v = 0, s = 0.1 m → t = √ s / 0.5 a = √ 0.1 / 0.5 * 9 * 10¹⁵ = 4.7 * 10^-9 s.
4. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (4.7 * 10^-9)² = 0.1 m s^-1.

5. x = v t → v = x / t = 0.1 / 4.7 * 10^-9 = 21 276 596 m s^-1.

6. t = √ s / 0.5 a = √ 0.02 / 0.5 * 9 * 10¹⁵ = 2.1 * 10^-9 s.
7. y = 0.5 (e / m) * E * t² = 0.5 * 1.8 * 10¹¹ * 50 000 * (2.1 * 10^-9)² = 0.02 m s^-1.

8. v = x / t = 0.02 / 2.1 * 10^-9 = 9 523 809 m s^-1.

Check your value for s in line 3. The plate separation was given in mm, not cm.

I don't know what lines 4 through 6 were meant to accomplish. Line 3 was meant to find the transit time through the plates.

If you fix up the calculation in line 3 (and keep more figures in intermediate values! Don't truncate or round values in intermediate steps except for presentation purposes; never use rounded values to continue calculations) then you should be almost home and done.

You would also do well to heed what @PeroK said about using symbols rather than hastening to plug in numbers. Numbers are error sensitive, prone to typos, and it's easy to forget what they represent as you carry them through multiple steps. With symbols you always know exactly what you are dealing with.