11-71. A particular proton in space is subjected to two fields: a gravitational field (1.30X105 N/kg  toward planet X), and an electric field (2.47X10-4 N/C away from planet X).
 (a) If the proton is released from rest, in which direction will it move?
 (b) After moving 1.50 m, what will be its speed? Use energy methods and assume uniform fields.

ACCUMULATED SOLUTION

FE = qE     Fg =  mg
Fg =  mg = (1.673X10-27 )(1.3X105) = 2.175X10-22 N
FE = qE = (1.602X10-19 )(2.47X10-4) = 3.957X10-23 N


Correct.
So if  EK is kinetic energy, EP is gravitational potential energy, and U is the Electric Potential energy we have for the conservation of energy :

EK1 +EP1 +U1 = EK2 +EP2 +U2
and EK1 is equal to:

(A)   0
(B)   EK2
(C)   mg