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anonymous
 one year ago
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.
anonymous
 one year ago
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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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1devil might be in detail. what are you using for the electrical field of the sheet?

Nurali
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/4f54fc54e4b01ed6138379db

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0use Gauss' Law to obtain E above the sheet; is E up, or is E down? Is the Electric Force up, or down?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the situation described in your problem is: dw:1436512391791:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0of course, we can neglect the weight force of our pointlike 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: \[\Large \begin{gathered} q = 3 \times {10^{  6}}coulombs \hfill \\ \\ E = \frac{\sigma }{{{\varepsilon _0}}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here, 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: \[\Large v = \frac{{qE}}{m}t\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436513140003:dw

alekos
 one year ago
Best ResponseYou've already chosen the best response.1The 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 2.17E6 N Ft = Fg  Fe = mg  σq/ε

alekos
 one year ago
Best ResponseYou've already chosen the best response.1Then use v^2 = 2as to find the velocity where a = Ft/m and s = 0.60.25

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1+1 @Alekos agree apart from the field. => E = σq/2ε approach gives "sensible answers" too

alekos
 one year ago
Best ResponseYou've already chosen the best response.1Yes you're right! E = σ/2ε for a very large flat sheet. my mistake, thanks

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0oops..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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0of course those equation are compatible with your data, if: \[\Large \frac{{qE}}{m}  g < 0\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1of 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]: \(2 (\frac{1}{2})m \dot z \ddot z  \frac{q \sigma}{2 \epsilon} \dot z + mg \dot z = 0\) \(m \ddot z  \frac{q \sigma}{2 \epsilon} + mg = 0\) \( \ddot z = \frac{q \sigma}{2 \epsilon m}  g\) \( \dot z = v = [\frac{q \sigma}{2 \epsilon m}  g] t\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1forgive any typo's
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