Question

question

Answer 1

\[\Delta y=\Delta y_1+\Delta y_2=\frac 12 a(\frac{L}{v_0})^2+L +D \]
\[x_1=v_0t\]
\[t=\frac{x_1}{v_0}=\frac{L}{v_0}\]
\[\Delta y_1=v_0t+\frac 12 at^2\]
\[\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2\]
\[\Delta y_1=\frac 12 a(\frac{L}{v_0})^2+L\]
now outside o the acceleration plates
\[x_2=v_0t\]
\[t=\frac{x_2}{v_0} =\frac D{v_0}\]
\[\Delta y_2=v_0(\frac D{v_0})\]

Here is where I'm stuck:
\[ma=qE\]
\[a=\frac {qE}{m}\]
\[KE=U\]
\[\frac 12 mv_0^2=qV_acc\]
\[v_0^2=\frac {2qV_acc}{m}\]

now i Have
\[\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D\]

Answer 2

but my equation should look like
\[\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)\]

Answer 3

I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess

Answer 4

feel free to write on here fellas:)

Answer 5

\[y_1=v_yt+\frac 12 at^2\]
is there a y velocity within the plates?

Answer 6

please stop posting @CaritaDeAngel

Answer 7

can I type today on OS (testing)

Answer 8

please continue posting @TuringTest

Answer 9

Okay let's start with the electron gun ...