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.

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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.

Physics
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devil might be in detail. what are you using for the electrical field of the sheet?
use Gauss' Law to obtain E above the sheet; is E up, or is E down? Is the Electric Force up, or down?

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the situation described in your problem is: |dw:1436512391791:dw|
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: \[\Large \begin{gathered} q = 3 \times {10^{ - 6}}coulombs \hfill \\ \\ E = \frac{\sigma }{{{\varepsilon _0}}} \hfill \\ \end{gathered} \]
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: \[\Large v = \frac{{qE}}{m}t\]
|dw:1436513140003:dw|
The force due to gravity is greater than the force due to the electric field so it will accelerate towards the sheet with a total force of 2.17E-6 N Ft = Fg - Fe = mg - σq/ε
Then use v^2 = 2as to find the velocity where a = Ft/m and s = 0.6-0.25
+1 @Alekos agree apart from the field. => E = σq/2ε approach gives "sensible answers" too
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Yes you're right! E = σ/2ε for a very large flat sheet. my mistake, thanks
oops..the weight force is not neglectable, so the right equation of motion are: \[\Large \begin{gathered} v = \left( {\frac{{qE}}{m} - g} \right)t \hfill \\ \\ z = {z_0} + \frac{1}{2}\left( {\frac{{qE}}{m} - g} \right){t^2} \hfill \\ \end{gathered} \] where g is the earth gravity |dw:1436545462855:dw|
of course those equation are compatible with your data, if: \[\Large \frac{{qE}}{m} - g < 0\]
of course, the OP specified "Using energy methods" so instead, and using @Michele_Laino 's handy raference system [z points upwards], we can say: \(\frac{1}{2}m \dot z^2 - \frac{q \sigma}{2 \epsilon}z + mgz = const\) "if" the sphere moves upwards or downwards, it increases its KE. if it goes up [z increases], it loses electrical PE [as two repelling +ve charges are further apart] but gains gravitational potential [as 2 "attractive" masses are further apart]. And vice versa. we can then differentiate this wrt t [and we should go back to the force equations]: \(2 (\frac{1}{2})m \dot z \ddot z - \frac{q \sigma}{2 \epsilon} \dot z + mg \dot z = 0\) \(m \ddot z - \frac{q \sigma}{2 \epsilon} + mg = 0\) \( \ddot z = \frac{q \sigma}{2 \epsilon m} - g\) \( \dot z = v = [\frac{q \sigma}{2 \epsilon m} - g] t\)
forgive any typo's

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