The acceleration of an electron in the electric field of a positively charged sp

Question

The acceleration of an electron in the electric field of a positively charged sphere is inversely proportional to the square of the distance between the electron and the center of the sphere. Let an electron fall from rest at infinity to the sphere. What's the electron's velocity when it reaches the surface of the sphere? The answer to this problem is v=v(2k/ma). However, I do not know how to get it. Please help. Thank you.
The acceleration of an electron in the electric field of a positively charged sphere is inversely proportional to the square of the distance between the electron and the center of the sphere. Let an electron fall from rest at infinity to the sphere. What's the electron's velocity when it reaches the surface of the sphere? The answer to this problem is v=v(2k/ma). However, I do not know how to get it. Please help. Thank you.

Explanation / Answer

from the answer we know acceleration = -k/(mr) where m is mass, r is the distance between electron and the center of sphere. dv/dt = -k/(mr^2) dv/dr * dr/dt = -k/(mr^2) v dv/dr = -k/(mr^2) mv dv = -k/r^2 dr integrate mv^2/2 = k/r + C when r --> infinity, v = 0, so C = 0 v^2 = 2k/(mr) when r = a (radius of sphere) v^2 = 2k/(ma) so v = sqrt[2k/(ma)]

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