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pmb_phy
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Its funny how one can forget how to do very simple things.
Here is a problem I've been working on and have a mental block and can't get the darn solution. It's from Krauss's EM text
First thing to do is to find vx which is found from the kinetic energy which it gains as it accelerates through a difference of potential Va. Therefore
[tex]K = \frac{1}{2}mv_x^2 = eV_a[/tex]
solve for [tex]mv_x^2[/tex] to get
[tex]v_x = \sqrt{\frac{2eV}{m}}[/tex]
The y-component of velocity is found through
[tex]F = ma_y = eE_y = eV_d/d[/tex]
[tex]a_y = \frac{eV_d}{md}[/tex]
[tex]y = a_y t^2 = \frac{eV_d}{md}t^2[/tex]
where t is the time it takes the particle to travel a distance l. So t = l/vx. Upon substitution this gives
[tex]y = a_y t^2 = \frac{eV_d}{md}\frac{l^2}{(ev_x)^2}[/tex]
[tex]y = \frac{V_d}{d}\frac{l^2}{2V_a}[/tex]
[tex]y = \frac{V_d l^2}{2V_ad}[/tex]
That's about as far as I've gotten. Unless they were thinking that one should find v_y/v_x and note that v_y/v_x = y/x??
Pete
Here is a problem I've been working on and have a mental block and can't get the darn solution. It's from Krauss's EM text
Note: Assume non-relativistic motion. I'm assuming beam enters between places with d/2 on each side(page 206 Problem 5-4-12)
Cathode-ray tube. Electric field deflection Show that thte deflection of an electron at the screen of a cathode-ray tube is given by
[tex]y = \frac{V_d l x}{2V_a d}[/tex]
where
y = deflection distance at screen, m
Vd = deflecting potential, V
L = length of deflecting plate, m
x = distance from deflecting plates to screen, m
Va = accelerating potential, V
d = spacing of deflecting plates, m
First thing to do is to find vx which is found from the kinetic energy which it gains as it accelerates through a difference of potential Va. Therefore
[tex]K = \frac{1}{2}mv_x^2 = eV_a[/tex]
solve for [tex]mv_x^2[/tex] to get
[tex]v_x = \sqrt{\frac{2eV}{m}}[/tex]
The y-component of velocity is found through
[tex]F = ma_y = eE_y = eV_d/d[/tex]
[tex]a_y = \frac{eV_d}{md}[/tex]
[tex]y = a_y t^2 = \frac{eV_d}{md}t^2[/tex]
where t is the time it takes the particle to travel a distance l. So t = l/vx. Upon substitution this gives
[tex]y = a_y t^2 = \frac{eV_d}{md}\frac{l^2}{(ev_x)^2}[/tex]
[tex]y = \frac{V_d}{d}\frac{l^2}{2V_a}[/tex]
[tex]y = \frac{V_d l^2}{2V_ad}[/tex]
That's about as far as I've gotten. Unless they were thinking that one should find v_y/v_x and note that v_y/v_x = y/x??
Pete
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