A large number of energetic cosmic-ray particles reach Earth\'s atmosphere conti

Question

A large number of energetic cosmic-ray particles reach Earth's atmosphere continuously and knock electrons out of the molecules in the air. Once an electron is released, it responds to an electrostatic force that is due to an electric field E produced in the atmosphere by other point charges. Near the surface of Earth this electric field has a magnitude of |E| = 150 N/C and is directed downward, as shown in the figure below. Calculate the change in electric potential energy of a released electron when it moves vertically upward through a distance d = 703 m.

Explanation / Answer

Hi,

In this case, we can put the electric field as a vector over the y axis in a coordinate system (x,y). Then we should remember the relation between the electric field and the electric potential, as well as the relation between the last and the electric potential energy:

Ey = -dV / dy (1); where Ey is the component of the electric field over the y axis, and V is the electric potential.

U = qV (2); where U is the change in the electric potential energy while V is the change in the electric potential.

If we integrate (1) we get the following:   V = -Ey * d ; where d is the distance traveled by the electron, which is positive because is increasing the value of y (the electron is moving upward).

As we used the traditional coordinate system, the component of the electric field over the y axis (which is this case is the entire electric field) is negative as well.

Ey = 150 (-j) , by this reason we have the following:

V = (150 N/C)*(703 m) = 106*103 V

Now we can find the change in electric potential energy:

U = qV = (-1.602*10-19 C)*(106*103 V) = -1.698*10-14 J

I hope it helps.

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