11-71. A particular proton in space is subjected to two fields: a
gravitational field (1.30X10^{5} 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

*F*_{E} = qE
F_{g} = mg

*F*_{g} = mg
= (1.673X10^{-27} )(1.3X10^{5}) = 2.175X10^{-22}
N

*F*_{E} = qE = (1.602X10^{-19}
)(2.47X10^{-4}) = 3.957X10^{-23} N

Correct.

So if *E*_{K} is kinetic
energy,
*E*_{P }is gravitational potential energy, and *U*
is
the Electric Potential energy we have for the conservation of energy :
*E*_{K1} +*E*_{P1 }+*U*_{1
}=
*E*_{K2}
+*E*_{P2} +*U*_{2}

and *E*_{K1} is equal to:

(A) 0

(B) *E*_{K2}

(C) *mg*