The electric field near the surface of the earth points downward and has a magnitude of 150 N/C.
(a) Compare the upward electric force on an electron with the downward gravitational force.
(b) What charge should be placed on a penny of mass 3 g so that the electric force balances the weight of the penny near the earth's surface?
2007-05-01 20:05:10 · 2 answers · asked by hpage 3 in Science & Mathematics ➔ Physics
The electric force is q*E; for and electron, q = -1.6*10^-19 coul. Therefore for a field of 150N/coul, the force is -1.6*10^-19 * 150 N = 2.4*10^-17N (upward) The gravitational force is m*g, where m is the mass of the electron = 9.1*10^-31 kg. g = 9.8m/sec^2 = 8.92*10^-30N. The ratio of electric force to gravity is 0.269*10^14
The weight of the penny is 0.003*g = 0.029N. q = F/E, q = 0.029/150 = 1.93*10^-4 coul
2007-05-01 21:03:36 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
electric powered container E= - dV /dr As ability is continuous dV / dr =0 for this reason electric powered container is 0 in a area the place electric powered ability is continuous. for example, interior a hollow charged matallic sphere, the electrical powered container IS 0 however the aptitude interior is continuous and equivalent to ability on the suface of charged sphere
2016-10-14 08:09:36 · answer #2 · answered by ? 4 · 0⤊ 0⤋