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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
tdl17
2019-01-05 22:45:55		You can also use dimensional analysis. Whatever that goes into the arc tangent should be purely numerical without any units, and the only answer where this happens is choice A. Vq cancels with mv^2 because they both have units of Joule. L cancels with d because they both have units of length. NEC
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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You can also use dimensional analysis. Whatever that goes into the arc tangent should be purely numerical without any units, and the only answer where this happens is choice A. Vq cancels with mv^2 because they both have units of Joule. L cancels with d because they both have units of length.

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