Conservation of energy and electrons in electric field?

[okgo]

Homework Statement

http://www.screencast.com/users/trinhn812/folders/Jing/media/6596c399-feb6-49cf-82a7-d6d16c2a88e8
Question 17

Homework Equations

So I solved this problem using kinematics and would like to learn how to solve it use conservation of energy. I can't seem to get the right answer.

I did qEd=.5mv^2

The Attempt at a Solution

m=9.11E-31
q=1.60E-19
E=6000 (I calculated by 60/.01)
d=?? ( If I label it as .01m the answer is wrong.)

Correct answer is marked as 7E6 m/s for velocity in y-direction

 

[kuruman]

It seems you are forgetting that the kinetic energy is the sum of two terms

KE = (1/2)m(v_x² + v_y²)

 

[AndyNewton]

Using energy to calculate the vertical speed is an interesting idea. It would be an elegant method of finding the vertical speed if the question had given us a value for the vertical displacement of the electron beam between the point where it enters the gap between the plates and the point where it exits the gap.

If that displacement was known (though in the question it is not given) as a vertical displacement = J, then:

Work done by the field acting on the electron = q.E.J

Change in kinetic energy of electron = (1/2).m.(Vy^2) where Vy is vertical componet of electron velocity.

And since work done = change in kinetic energy we can write,
q.E.J = (1/2).m.(Vy^2)
and then rearrange the equation to give
Vy = sqroot ( 2.q.E.J / m ).

Unfortunately, in this question we are not given a value for J. Despite that, an energy calculation is still possible of course, but it would just be a longer more roundabout variation of the more direct calculation method using force, time, acceleration etc, as already calculated.

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Considering kuruman's comment above, it would in fact be possible to ignore the horizontal component of the electron velocity in an energy calculation.
If you were expected to "show all working" in an examination question then you would probably have to demonstrate that fact also in a full answer as shown below, making the energy method even longer to work through:

kuruman is correct in stating that the total kinetic energy of the electron is
KE = (1/2).m.V^2 = (1/2)m(Vx^2 + Vy^2)

so the change in kinetic energy is therefore
delta(KE) = (1/2).m.delta(V^2) = (1/2).m.(delta(Vx^2) + delta(Vy^2))
where delta(...) means "the change in the (...) value".

As Vx is constant we know that delta(Vx^2) = 0
As Vy is initially zero we know that delta(Vy^2) = (final value of Vy ^ 2)

So delta (KE) = (1/2).m.(Vy^2)
... and that's nice to know, even though it did not help solve this particular exam question :-)
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[okgo]

sweet thanks a lot. what a tricky question