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anonymous

  • one year ago

A projectile has a mass m and, of course, it is charged (q). The tube, has an electric field E as shown which deflects the projectile as shown. The mass enters the tube with a velocity v (tube length is L) and, due to the electric field, it will be deflected in an upward direction. For a projectile mass of 1.0x10^-10kg, an electric field strength of 1.0x10^6N/C, a plate length of 15mm, and a drop initial horizontal speed of 29m/s, calculate the drop charge necessary to result in a deflection of 1.3mm when it first emerges from the tube.

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  1. anonymous
    • one year ago
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    |dw:1441757085723:dw| Here is the diagram that I attempted to draw

  2. anonymous
    • one year ago
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    @Mertsj @texaschic101 @Nnesha @Luigi0210 @robtobey Would any of you be able to help me, please? :)

  3. anonymous
    • one year ago
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    @e.mccormick @UnkleRhaukus

  4. UnkleRhaukus
    • one year ago
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    F = qE is the force of the electric field on the charge. Are there any other forces acting?

  5. Jhannybean
    • one year ago
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    How about taking into consideration the difference in height when the charge moves across the field? \(\sf \Delta y = v_{i_x}t+\frac{1}{2}gt^2\)

  6. Michele_Laino
    • one year ago
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    the equation of motions are: \[\Large \left\{ \begin{gathered} z\left( t \right) = - \frac{{\left| q \right|E}}{{2m}}{t^2} + {h_0} \hfill \\ \hfill \\ x\left( t \right) = {v_0}t \hfill \\ \end{gathered} \right.\] |dw:1441829580320:dw|

  7. Michele_Laino
    • one year ago
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    now the projectile emerges from tube at time \( \large \tau \), such that: \[\Large x\left( \tau \right) = d = {v_0}\tau \to \tau = \frac{d}{{{v_0}}}\] at that time the z-coordinate is: \[\Large z\left( \tau \right) = - \frac{{\left| q \right|E}}{{2m}}{\tau ^2} + {h_0}\] |dw:1441829944559:dw|

  8. Michele_Laino
    • one year ago
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    so, we can write this: \[\Large \begin{gathered} \delta = {h_0} - z\left( \tau \right) = \frac{{\left| q \right|E}}{{2m}}{\tau ^2} \hfill \\ \hfill \\ \delta = \frac{{\left| q \right|E}}{{2m}}\frac{{{d^2}}}{{v_0^2}} \hfill \\ \end{gathered} \] where \( \large \delta \) is the deflection

  9. Jhannybean
    • one year ago
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    Just wondering... but why are you using Tau?

  10. Michele_Laino
    • one year ago
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    usually, when I write an equation, I use "t" for as variable, whereas I use "\tau" to denote a specific time @Jhannybean

  11. Jhannybean
    • one year ago
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    Oh, interesting!

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