Electric field

  • #1
ok i have a problem to work on in my new course, and i was wondering what i need to do to tackle it. the question is as follows:

An electron is projected with an initial velocity Vo=10^7m/s into the uniform field between the parallel plates "E". the direction of the field is vertically downward and the field is zero everywhere else except between plates. the electron enters the field at a point midway the plates. if the electron just misses the upper plate as it uses energy from the field, find the magnitude of the field "E"

are these the formulas needed
F = ma = qE
s = ut + 0.5 at^2
v = u + at
v^2 = u^2 + 2as

and what does ( a, u, and s) stand for?
 
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Answers and Replies

  • #2
Hello there,
Yes, you've selected the proper formulae, but the questions you posed pertain to a far greater lack of basic physical understanding; So perhaps reading up on the more elementary concepts of mechanics could lead you forward.
Respectively, and as is typical in physical notation, a stands for acceleration, or [itex] \Large \frac{d^2\vec{r}}{dt^2} [/itex], the Second derivative of the displacement, due to time.
u in your case you've selected as some [itex] \large v_0 [/itex] or an initial velocity of some kind.
S - is the displacement.

Daniel
 
  • #3
ok since im given initial velocity and displacement which is the length of the plate (2cm) then how do i find the acceleration
 
  • #4
Okay, in order to step further into this, you'd have to imagine what forces are at play here.
On the one hand, your initial velocity is oriented, parrallel to the plates, but the force is perpendicular.
Lets first create our axis system: Y is the axis along which E exists, and X is the axis perpendicular to it.
Putting it together: [itex] \vec{E} || \hat{y} [/itex], [itex] \vec{v_0} || \hat{x} [/itex]
Now, is there gravity at play?
We're missing data on the moment of exit, what's the angle there, other information?
Please provide everything given first...
Daniel
 
  • #5
ok i beleive gravity is in play here, attached is diagram i scanned
 

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  • #6
Due to the statement of your problem, unless you had omitted something, this is going to get slightly messy :):
So, firstly, there's no change in the absolute value of the velocity, since the electric field acts perpendicular to the initial velocity.
this means that at any moment in time [itex] |\vec{v}| = v_0 [/itex]
v_y which had no initial velocity in it, we'll define as:
[itex] v_y = at -> a = a_e-g [/itex] a_e is the acceleration due to the electric field.
Integrating, that gives us: [itex] Y = {at^2}/2 [/itex]
We need to find time t' when, the particle has finished travelling 1 cm, which I'll call h.
that gives us [itex] t' = \sqrt{\frac{2h}{a}} [/itex]
And now x, which is problematic, somewhat, and requires greater manipulation:
Since |v| is constant(see above), we'll get [itex] |v| = \sqrt{{v_x}^2+{v_y}^2} [/itex], or [itex] v_x = \sqrt{v^2-{v_y}^2} [/itex], or by substituting v_y, [itex] v_x = \sqrt{v^2-{(at)}^2} [/itex], and x is the [itex] x = \int_0^t' \sqrt{v^2-{(at)}^2}dt [/itex], and by replacing t = (v/a)*sin(u), we get dt = vcos(u)*du/a and so forth.
This results in:
x(t') = d(2 cm)= [itex] \Large \frac{at'\sqrt{v^2 -(a^2t'^2)}+v^2\arctan(\frac{at'}{\sqrt{v^2-a^2t'^2}})}{2a}
[/itex]
Plug-in t' above and find a...
Daniel
 
  • #7
lol it sure got messy, but i understand most of it. the only questions i have are: with v_x do you mean velocity with respet to x, and what do you mean by {itex}
 
  • #8
Yes, v_x is the velocity parrallel to x, and itex is used for the Latex format on this forum...
It really helps with the illustrations,
Daniel
P.S
I am in the process on calculating it myself now, so within the next few minutes I should have "a" here...
Feel free to query/question any segment you don't get, I'll gladly elucidate matters...
Are you sure's there's no further information given? What's q?
 
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  • #9
Still waiting on an answer here :)...
 
  • #10
gneill
Mentor
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So, firstly, there's no change in the absolute value of the velocity, since the electric field acts perpendicular to the initial velocity.
That's not true. There will be no change in the horizontal component of the velocity, but the vertical component will change, thus the speed will change. This is just like a projectile under the influence of a constant gravitational field, only in this case it's an electric field providing the force (and hence acceleration).
 
  • #11
You're absolutely right, and I do wonder how did I end up with the fixation that somehow the total velocity was to be preserved; Tough day I guess, or my attempt to convolute and complicate matters...
Thank you for pointing out such an egregious error,
This does simplify deriviation...
Sorry for the mixup then,
Daniel
 
  • #12
hey danielakkerma after staring at the free body diagram for a while i finally came to the conclusion that gneill is absolutely right! and that this is very much like a projectile with an initial velocity which is influenced by an acceleration mimicking that of gravity. also since the angle at which it is projected is zero, it actually makes things easier! i basically set up these following equations [itex]\Delta[/itex]X=V0t and
[itex]\Delta[/itex]Y=(1/2)at2. we solve for t in the deltaX equation substitute it for t in the deltaY equation. solve for acceleration. then we have F=ma. we use mass*acceleration to find the force. then we fave F=-eE. after some algebraic manipulation we come to find that E=(ma/-e)!! man it feels so good to get it right!! thanks for all your help you guys.
 

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