Calculate Speed of Electron from Charge Density

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The speed of an electron released from rest above a uniformly charged infinite plane can be calculated using the Work Energy theorem. The relevant equation is v = √(2kqQ/mr), where q is the electron's charge, Q is the charge density of the plane, and m is the electron's mass. For a charge density of 10^-9 C/m² and the electron's charge and mass, the calculated speed upon hitting the plane is approximately 1.9 x 10^6 m/s. This approach effectively applies the principles of potential and kinetic energy in the context of a continuous charge distribution. The discussion highlights the transition from point charge calculations to those involving charge density.
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** Edit: Nevermind. I figured it out using the Work Energy theorem.

An electron is released from rest 1.0 cim above a uniformly charged infinite plane with a charge density of 10-9C/m2. What is the speed of the electron when it hits the plane?

my attempt:
Potential energy when it is released= kinetic energy when it hits.

kqQ/r = 0.5 mv2

isolate v:

<br /> v = \sqrt {\frac{{2kqQ}}{{m \cdot r}}} <br />

This would work if I was given 2 point charges, but how do I do this with a charge density and a point charge?


 
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Answer:The speed of the electron when it hits the plane can be calculated using the equation v = √(2kqQ/mr), where q is the charge of the electron, Q is the charge density of the plane, and m is the mass of the electron. In this case, q=1.6x10-19 C, Q=10-9 C/m2, and m=9.1x10-31 kg. Plugging in these values gives a speed of 1.9x106 m/s.
 

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