Question

A small sphere with mass 5.00.×10−7kg and charge +3.00μC is released from rest a distance of 0.600 m above a large horizontal insulating sheet of charge that has uniform surface charge density σ=+8.00pC/m2.

a)Using energy methods, calculate the speed of the sphere when it is 0.250 m above the sheet of charge?

I just need help understanding what I have to do to get the speed. I tried finding the difference in potential to calculate the Work done and then find the final velocity from that. I tried using conservation of energy using Electric Potential energy, gravitational PE and KE.

Answer 1

devil might be in detail.

what are you using for the electrical field of the sheet?

Answer 2

http://openstudy.com/study#/updates/4f54fc54e4b01ed6138379db

Answer 3

use Gauss' Law to obtain E above the sheet; is E up, or is E down?
Is the Electric Force up, or down?

Answer 4

the situation described in your problem is:

Answer 5

of course, we can neglect the weight force of our point-like charge. Furthermore, we have the subsequent equations:

\[\Large \begin{gathered}
v = \frac{{qE}}{m}t \hfill \\
\hfill \\
z = {z_0} + \frac{1}{2}\frac{{qE}}{m}{t^2} \hfill \\
\end{gathered} \]
where:
\[\Large \begin{gathered}
q = 3 \times {10^{ - 6}}coulombs \hfill \\ \\
E = \frac{\sigma }{{{\varepsilon _0}}} \hfill \\
\end{gathered} \]

Answer 6

here, you have to solve the second equation for t, namley:
\[\Large t = \sqrt {\frac{{2m}}{{qE}}\left( {z - {z_0}} \right)} \]
where z=0.250 meters
then you have to substitute the resultant value, into the first equation:
\[\Large v = \frac{{qE}}{m}t\]