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#1
Its part d thats confusing me; I'll have it attached as an image. I have working out the acceleration correctly to be 1.76*10^15 but Im not sure how to do the next part using this?? Any help is greatly appreciated thanks.

Edit: Not letting me post? :/

But the question basically says, "Electrons are accelerated upwards by the electric force on them. Calculate their acceleration, and use your answer to deduce the upward component of the electron's velocity as it emerges from the space between the plates."

So I've managed to work out the acceleration correctly but how would I use this for the second part its asking for?? Im so confused
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2 years ago
#2
Equations of motion?

Without seeing the Q I'm fumbling in the dark, but I'm imagining the electrons enter some kind of electric field which is at right angles to their direction of motion?

If that's the case, the you will need to use the horizontal speed to work out how long they will be between the plates, then use this time, plus the acceleration, to work out what their upwards/vertical velocity will increase to in this time.

Remember it's like projectile motion so horizontal and vertical motion are completely independent which means that, assuming they enter the plates horizontally, their vertical velocity at that time is zero.

Of course, your Q might be nothing like that, but it's my best guess until I see it....
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#3
(Original post by phys981)
Equations of motion?

Without seeing the Q I'm fumbling in the dark, but I'm imagining the electrons enter some kind of electric field which is at right angles to their direction of motion?

If that's the case, the you will need to use the horizontal speed to work out how long they will be between the plates, then use this time, plus the acceleration, to work out what their upwards/vertical velocity will increase to in this time.

Remember it's like projectile motion so horizontal and vertical motion are completely independent which means that, assuming they enter the plates horizontally, their vertical velocity at that time is zero.

Of course, your Q might be nothing like that, but it's my best guess until I see it....
I think it'll let me post the image in a reply.. does this help? It is part d and the figure 9.27 is shown below
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2 years ago
#4
Yes, as I imagined, so try doing as suggested in my first reply.

How long will it take the electron to travel the 10cm horizontally and how much will its vertical speed increase in this time.

(A variation on this question is to ask about its vertical displacement in the time taken, in order to work out whether it emerges from the plates or hits one of them first.)
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#5
(Original post by phys981)
Yes, as I imagined, so try doing as suggested in my first reply.

How long will it take the electron to travel the 10cm horizontally and how much will its vertical speed increase in this time.

(A variation on this question is to ask about its vertical displacement in the time taken, in order to work out whether it emerges from the plates or hits one of them first.)
Alright thank you so much. There is another question here again just part d - the very last part Im stuck on. Could you please explain it;

The answer says the magnetic field will be much greater than the electric field and so will curve the ion upwards - but I don't understand why this is so - why'd the magnetic field greater?
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#6
(Original post by phys981)
Equations of motion?

Without seeing the Q I'm fumbling in the dark, but I'm imagining the electrons enter some kind of electric field which is at right angles to their direction of motion?

If that's the case, the you will need to use the horizontal speed to work out how long they will be between the plates, then use this time, plus the acceleration, to work out what their upwards/vertical velocity will increase to in this time.

Remember it's like projectile motion so horizontal and vertical motion are completely independent which means that, assuming they enter the plates horizontally, their vertical velocity at that time is zero.

Of course, your Q might be nothing like that, but it's my best guess until I see it....
The acceleration I'd use here would be 9.81 right? as it vertical motion? So I don't use the acceleration I just worked out? or do i...

EDIT: I see that I get the wrong answer using 9.81 and the right answer using the acceleration I worked out - why is this the case? that acceleration I worked out - is it just for the vertical direction since it says going UPWARDS
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2 years ago
#7
There are now two forces acting on those charges, in opposite directions

E-field will exert force Eq in one direction, B-field will exert force Bqv in the other.

For the charges to pass through the slit, they must be undeflected which means the resultant force on them must be zero.

Hopefully that's enough for you to answer it.
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2 years ago
#8
(Original post by MrToodles4)
The acceleration I'd use here would be 9.81 right? as it vertical motion? So I don't use the acceleration I just worked out? or do i...

NOOOOOOO!!!!!! Use the one you've worked out.

Gravitational pull on an electron will be much smaller than the force due the the E field so ignore it.
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#9
(Original post by phys981)
NOOOOOOO!!!!!! Use the one you've worked out.

Gravitational pull on an electron will be much smaller than the force due the the E field so ignore it.
Sorry, I got it now thank you so much for all your help honestly.
1
#10
(Original post by phys981)
There are now two forces acting on those charges, in opposite directions

E-field will exert force Eq in one direction, B-field will exert force Bqv in the other.

For the charges to pass through the slit, they must be undeflected which means the resultant force on them must be zero.

Hopefully that's enough for you to answer it.
I understand that it won't be equal anymore - so magnetic force is greater as BQV and speed is greater. So magnetic force always acts towards the positive? i.e. upwards here? and so the electron will curve upwards... and the electric fields act downwards towards the negative plate?
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2 years ago
#11
(Original post by MrToodles4)
I understand that it won't be equal anymore - so magnetic force is greater as BQV and speed is greater.
yes, so it will accelerate (vertically) under what is now an unbalanced force, which means it will be deflected and will miss the slit.
1
#12
(Original post by phys981)
yes, so it will accelerate (vertically) under what is now an unbalanced force, which means it will be deflected and will miss the slit.
The last question I have on this particular question is regarding the forces - which direction do the magnetic force and electric force act? would I use left hand rule or right hand rule or something or is it conventional? Sorry for all these
0
2 years ago
#13
The Q tells us that they are + charges so we know they will be deflected downwards by the electric field, ie towards the negative plate.

I would (lazily) assume that the magnetic field is in the opposite direction - we know this has to be true if some pass through undeflected. However, if you want to check, it's the Left hand rule for magnetic force, yes.
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#14
(Original post by phys981)
The Q tells us that they are + charges so we know they will be deflected downwards by the electric field, ie towards the negative plate.

I would (lazily) assume that the magnetic field is in the opposite direction - we know this has to be true if some pass through undeflected. However, if you want to check, it's the Left hand rule for magnetic force, yes.
Okay makes sense We treat the ions as conventional current right? so they're going left to right. If they were negative ions would we still assume they go left to right in this question - because I know if they were electrons we'd take the opposite direction for conventional current.
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2 years ago
#15
Yes, they're positive so like conventional current.

And yes, with negative charges you have to watch what you do with the current finger in the L H rule.
0
#16
(Original post by phys981)
Yes, they're positive so like conventional current.

And yes, with negative charges you have to watch what you do with the current finger in the L H rule.
You know in that first question of this thread you helped me with d, Im stuck on e sorry... Like Im thinking I use f=BQVsinx - is this right? but I don't know B
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2 years ago
#17
This is just trig. You know the horizontal and vertical components of the velocity - draw them as vectors and find the angle between.

something like this

You probably did some stuff like this when you did the forces and motion topic, but that will seem like a long time ago now.
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#18
(Original post by phys981)
This is just trig. You know the horizontal and vertical components of the velocity - draw them as vectors and find the angle between.

something like this

You probably did some stuff like this when you did the forces and motion topic, but that will seem like a long time ago now.
OMG YES I remember wowww. Sorry about this, I would give a lot more reps if I could, You've helped me tonnes.

for part f I would use SUVAT right? I tried doing it with the information I already worked but I keep getting 2.5cm whereas the correct answer is 25cm. So I think I'm meant to look at it horizontally again - find the time, then use SUVAT again with the vertical motion to find the displacement? So horizontally would I use the distance from just outside the plates to the screen? or the whole distance of the entire diagram
0
2 years ago
#19
I will try it to check, 25cm is quite big for one of those screens...
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#20
(Original post by phys981)
I will try it to check, 25cm is quite big for one of those screens...
I was thinking that, thank you so much Ill be looking at g and onwards for now, I wish these things showed methods ergh. It so annoying how they just have answers
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