will rate fast - thank you for your time! There is an electric field of magnitud
ID: 2152010 • Letter: W
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will rate fast - thank you for your time!
There is an electric field of magnitude 150. N/C directed everywhere downward, near the surface of the Earth. What is the net electric charge on the Earth? You treat the Earth as a spherical conductor of radius 6371. km. What is the electrostatic potential at the Earth's surface, if the potential is taken to be zero at infinity? Using whatever data you need from Problem 1 above, and looking up any other necessary values: Calculate the acceleration-magnitude and direction-of a proton released near the surface of the Earth. Disregard any interaction with air molecules. Also, Calculate the charge-to-mass ratio of a particle which would hover in place if released near the surface of the Earth. Again, disregard any interaction with air molecules.Explanation / Answer
The weight of the person would be m*g, where m is the person's mass and g is the acceleration of gravity. The direction of the electrical field indicates that the earth's charge is negative, meaning that the charge on the person must also be negative for the two to repel. If your -153 N/C electric field is given as E, then the charge q is given by the formula F = qE. Solving this for q gives q = 56.0 kg * 9.81 m/sec / -153 N/C = -3.59 C Of course, the current from a sudden discharge would almost certainly electrocute the person if such a large charge were possible to accumulate in the first place. F = q1*q2 / (4*p*eo*r²) Here, q1 = q2 = -3.59 C, 1/(4*p*eo) = 8.99e9 N*m²/C², and r = 150 m, so F = q1*q2 / (4*p*eo*r²) = (-3.59 C)² * 8.99e9 N*m²/C² / (150 m)² = 5.15e6 N Being walloped with about five hundred tons of force is yet another reason that a person probably wouldn't survive acquiring such a charge.
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