Calculate the speed of the electrons as they enter the gap

[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 $v = 6.7 \times 10^7 ms^{-1}$