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Electromagnetism}Particle Trajectory

One can get a reasonable approximation for the deflection angle as follows.

Assuming that there is no magnetic field, one has from the Lorentz force F=ma=qE=qV/d, where one neglects gravitational acceleration. The acceleration is constant, and it is a=qV/(dm).

Recalling the baby physics kinematics equation, y=0.5at^2 \Rightarrow dy=atdt and the fact that x=L=vt \Rightarrow dx=vdt and t=L/v, one can calculate the angle as \tan \theta \approx dy/dx = \frac{atdt}{vdt}=at/v=\frac{qVL}{v^2dm}. Take the arctangent to get choice (A).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
Moush
2010-09-15 18:11:51
For a completely unphysical approach, see below. To actually learn something, see other posts.

I got this via simple guessing strategy.
(L/d) shows up in 3 answers; \(L/d)^2 and \(L/d)^{1/2} only once, eliminating C and E.
Vq shows up twice in the remaining answers; 2Vq once, eliminating D.
\(Vq/mv^2)^2 just looks wrong, eliminating B.
Alternate Solution - Unverified
Comments
deneb
2018-10-08 04:12:51
You have to divide both sides by v^2_x, and since v_x is just the original velocity given in the problem, that\\\\\\\'s how you get v^2 in the denominatorNEC
allenabishek
2017-08-03 19:29:24
Why is it that in the solution to find dy, the first term is ignored ?? the kinematic equation is actually y = v0t + 0.5at^2. why is the first term ignored?NEC
dipanshugupta
2017-03-22 14:15:07
Well A looked more beautiful. NEC
hooverbm
2012-11-02 09:15:49
Found a super fast way of doing this.

The charge is obviously acted upon by a force due to the electric field.

For parallel plates, the acceleration due to this force only acts along the y-direction. We know:

F = qE (V = Ed ---> E = V/d)

ma = qV/d

a = qV/md

The angle of deflection should be proportional to this term, leaving only answer A
NEC
soloeclipse
2010-11-04 16:31:02
All of the answers are dimensionless.
SonOfHam
2010-11-12 22:21:52
+1
kevintah
2015-10-12 19:03:36
lol, this sucks!
NEC
shjung84
2010-10-14 21:55:12
See dimension
The only possible choice is A of which argument indicates no dimension.
NEC
Moush
2010-09-15 18:11:51
For a completely unphysical approach, see below. To actually learn something, see other posts.

I got this via simple guessing strategy.
(L/d) shows up in 3 answers; \(L/d)^2 and \(L/d)^{1/2} only once, eliminating C and E.
Vq shows up twice in the remaining answers; 2Vq once, eliminating D.
\(Vq/mv^2)^2 just looks wrong, eliminating B.
ryanmes
2018-10-24 01:19:02
I did the same thing... it works
Alternate Solution - Unverified
dstahlke
2009-10-09 10:57:43
The diagram threw me off here... The deflection of the final velocity is arctan(dy/dx) but the dotted lines on the diagram form a triangle suggesting they want arctan(\Delta y / \Delta x) which I think would be equal to arctan(dy/2dx). That wasn't one of the options so in this case I suppose guessing the nearest answer would work.NEC
haro
2007-04-13 04:25:01
It is simplier to think of tan\theta = v_{y}/v_{x}
Jeremy
2007-10-15 12:01:19
I agree. From a=qV/dm use a=v_y/t, t=L/v_x, and finally, the equation you gave: \tan{\theta}=v_y/v_x.
walczyk
2011-04-05 13:27:12
this works well except i don't see where the \frac{1}{v^2} comes in?
NEC
yosun
2005-11-09 14:42:33
tachyon: thanks for the typo-alert; it has been corrected.NEC
tachyon
2005-11-09 13:36:53
typo-the correct answers is ANEC

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