How Do Free Electrons Move in Response to Atmospheric Potential Changes?

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Free electrons in the atmosphere move vertically upward in response to increasing atmospheric potential with elevation. As they rise, they experience a decrease in electric potential energy due to the potential gradient of 80 V per meter. To calculate the loss in electric potential energy for an electron moving 10 meters upward, the formula PE = qV is used, where q is the charge of the electron. The initial potential energy at 10 meters is calculated as -1.28 x 10^-17 J, while the final potential energy at the surface is 0 J. Thus, the total loss in potential energy for the electron is 1.28 x 10^-17 J.
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This question is giving me some difficulty.

Under normal atmospheric conditions, the potential in the atmosphere near the Earth's surface rises 80 V for every metre of increased elevation.
A) What direction will free electrons in the atmosphere move?
B) Calculate the loss in electric potential energy of a free electron which moves vertically through 10 m in the atmosphere.

For A) would the electrons move vertically since q1 and q2 are being moved further apart?
I don't know how to get the answer in B.
 
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A) I think they're asking if the electrons will move up or down, not just vertically or not. Do electrons seek lower or higher potential? Where will they find the potential they are looking for?
B) Do you know what the expression is for chagne in electric potential energy? ΔPele = qΔV. What's the change in potential then, if the particle moves 10 meters up or down? (You will know if it moves up or down from the answer to A).
 


A) Yes, the free electrons in the atmosphere would move vertically since the potential increases with elevation. As the electrons move higher, they would experience a decrease in potential energy.

B) To calculate the loss in electric potential energy, we can use the formula PE = qV, where PE is the potential energy, q is the charge of the electron, and V is the potential difference. In this case, q = -1.6 x 10^-19 C (the charge of an electron) and V = 80 V/m (the potential difference per meter).

So, the potential energy of the electron at the initial position would be PE1 = (-1.6 x 10^-19 C)(80 V/m)(10 m) = -1.28 x 10^-17 J.

At the final position, the potential energy would be PE2 = (-1.6 x 10^-19 C)(80 V/m)(0 m) = 0 J.

Therefore, the loss in potential energy would be ΔPE = PE2 - PE1 = 0 - (-1.28 x 10^-17 J) = 1.28 x 10^-17 J.

I hope this helps to clarify and solve the problem.
 
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