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GR9677 #71
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Electromagnetism$\Rightarrow$}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).

Alternate Solutions
 Moush2010-09-15 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.Reply to this comment
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.
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 denominator
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?
dipanshugupta
2017-03-22 14:15:07
Well A looked more beautiful.
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
soloeclipse
2010-11-04 16:31:02
All of the answers are dimensionless.
 SonOfHam2010-11-12 22:21:52 +1
 kevintah2015-10-12 19:03:36 lol, this sucks!
shjung84
2010-10-14 21:55:12
See dimension
The only possible choice is A of which argument indicates no dimension.
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.
 ryanmes2018-10-24 01:19:02 I did the same thing... it works
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.
haro
2007-04-13 04:25:01
It is simplier to think of tan$\theta$ = $v_{y}/v_{x}$
 Jeremy2007-10-15 12:01:19 I agree. From $a=qV/dm$ use $a=v_y/t$, $t=L/v_x$, and finally, the equation you gave: $\tan{\theta}=v_y/v_x$.
 walczyk2011-04-05 13:27:12 this works well except i don't see where the $\frac{1}{v^2}$ comes in?
yosun
2005-11-09 14:42:33
tachyon: thanks for the typo-alert; it has been corrected.
tachyon
2005-11-09 13:36:53

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.
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