## 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.

1. anonymous

|dw:1441757085723:dw| Here is the diagram that I attempted to draw

2. anonymous

@Mertsj @texaschic101 @Nnesha @Luigi0210 @robtobey Would any of you be able to help me, please? :)

3. anonymous

@e.mccormick @UnkleRhaukus

4. UnkleRhaukus

F = qE is the force of the electric field on the charge. Are there any other forces acting?

5. anonymous

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

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

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

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. anonymous

Just wondering... but why are you using Tau?

10. Michele_Laino

usually, when I write an equation, I use "t" for as variable, whereas I use "\tau" to denote a specific time @Jhannybean

11. anonymous

Oh, interesting!