Calculate the speed of the electrons as they enter the gap

[moenste]

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.

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 = 0.02 m → t = √ s / 0.5 a = √ 0.02 / 0.5 * 9 * 10¹⁵ = 2.108185107 * 10^-9 s.

Now we have electric field E, acceleration a and time t: t = √ d / (0.5 (V / d) * (e / m)) → t = d √ 2 m / V e.

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.

It gets the correct answer, but I have no idea how you derived it.

 

[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 = 0.02 m → t = √ s / 0.5 a = √ 0.02 / 0.5 * 9 * 10¹⁵ = 2.108185107 * 10^-9 s.

You've used the whole plate separation for the distance in line 3. The electron only traverses half that!

Really, you should start using symbols. You'd have caught that right away.

Regarding the derivation of PeroK's result, if you did your work using symbols then it would just be a matter of simplification of the expression at the end.

 

[moenste]

You've used the whole plate separation for the distance in line 3. The electron only traverses half that!

Really, you should start using symbols. You'd have caught that right away.

Regarding the derivation of PeroK's result, if you did your work using symbols then it would just be a matter of simplification of the expression at the end.

Using symbols:
E = V / d

a = E * (e / m) = (V / d) * (e / m)

s = v t + (1 / 2) a t² = v t + (1 / 2) (V / d) (e / m) t²

Doesn't look any easier.

 

[PeroK]

Using symbols:
E = V / d

a = E * (e / m) = (V / d) * (e / m)

s = v t + (1 / 2) a t² = v t + (1 / 2) (V / d) (e / m) t²

Doesn't look any easier.

This is for motion in the vertical, $y$, direction, where $v = 0$ (initial velocity in the y-direction) and $s = s_y$.

For what value of $s_y$ are you trying to find $t$?

 

[moenste]

This is for motion in the vertical, $y$, direction, where $v = 0$ (initial velocity in the y-direction) and $s = s_y$.

For what value of $s_y$ are you trying to find $t$?

s_y = v t + (1 / 2) a t² = v t + (1 / 2) (V / d) (e / m) t²

Since v = 0 then: s_y = (1 / 2) (V / d) (e / m) t².

s_y = 20 mm / 0.02 m.

 

[PeroK]

s_y = v t + (1 / 2) a t² = v t + (1 / 2) (V / d) (e / m) t²

Since v = 0 then: s_y = (1 / 2) (V / d) (e / m) t².

s_y = 20 mm / 0.02 m.

You're trying to find the time it takes the electron to be forced down from half way between the plates to the bottom plate. So, that's $s_y = d/2$, where $d$ is the distance between the plates. This gives:

$d/2 = (1/2)(V/d)(e/m)t^2$

Can you solve that for $t$?

 

[moenste]

You're trying to find the time it takes the electron to be forced down from half way between the plates to the bottom plate. So, that's $s_y = d/2$, where $d$ is the distance between the plates. This gives:

$d/2 = (1/2)(V/d)(e/m)t^2$

Can you solve that for $t$?

t = √ d² m / v e
t = d √ m / v e

 

[PeroK]

t = √ d² m / v e
t = d √ m / v e

You're nearly there! You need an equation for $u$ - the electron's horizontal motion - now. This would should be simpler.

 

[moenste]

You're nearly there! You need an equation for $u$ - the electron's horizontal motion - now. This would should be simpler.

x = v t?

 

[PeroK]

x = v t?

And what is $x$ in this case? How far does the electron travel horizontally in time $t$?

 

[moenste]

And what is $x$ in this case? How far does the electron travel horizontally in time $t$?

Horizontal component of velocity?

s = v t
where s = 5 d
v = 5 d / t?

 

[PeroK]

Horizontal component of velocity?

s = v t
where s = 5 d
v = 5 d / t?

Exactly. So, you just need to substitute the equation you have for $t$ now.

 

[moenste]

Exactly. So, you just need to substitute the equation you have for $t$ now.

t = 5 d / v

d √ m / V e = 5 d / v
d v √ m / V e = 5 d
v √ m / V e = 5
v = 5 / √ m / V e
v² = 25 V e / m
v = 5 √ V e / m = 5 √ 1000 * 1.8 * 10¹¹ = 67 082 039 m s^-1.

This should be correct.

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Thank you all!

 

[PeroK]

t = 5 d / v

d √ m / V e = 5 d / v
d v √ m / V e = 5 d
v √ m / V e = 5
v = 5 / √ m / V e
v² = 25 V e / m
v = 5 √ V e / m = 5 √ 1000 * 1.8 * 10¹¹ = 67 082 039 m s^-1.

This should be correct.

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Thank you all!

Yes, well done - you got there finally. Just one point. You had:

$vt = 5d$ and $t = d\sqrt{\frac{m}{Ve}}$

so:

$vd\sqrt{\frac{m}{Ve}} = 5d$

$v = 5\sqrt{\frac{Ve}{m}}$

was simpler.

Also, you should round your answer to $v = 6.7 \times 10^7 ms^{-1}$