Problem 11-71 Grav & Elec Potential - Part 3 - A

A particular proton in space is subjected to two fields: a gravitational field (\(1.30 \times 10^5\; N/kg\)  toward planet \(X\)), and an electric field (\(2.47 \times 10^{-4} \; N/C\) 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

Diagram B of proton subjected to two fields.

\(F_E = qE   \quad  F_g =  mg \\ F_g =  mg = (1.673 \times 10^{-27} )(1.3 \times 10^5) = 2.175 \times 10^{-22} \; N \\ F_E = qE = (1.602 \times10^{-19} )(2.47 \times 10^{-4}) = 3.957 \times 10^{-23} \; N \)

Correct

\(F_g > F_E\) so it moves toward the planet.

Diagram of proton moving toward planet.

Which statement is correct as the proton moves from position 1 to position 2?

(A)   Gravitational PE \((E_P)\) increases, Electrical PE \((U)\) decreases and KE increases

(B)   Gravitational PE \((E_P)\) decreases, Electrical PE \((U)\) increases and KE increases

(C)   Gravitational PE \((E_P)\) decreases, Electrical PE \((U)\) decreases and KE decreases

(D)   Gravitational PE \((E_P)\) increases, Electrical PE \((U)\) increases and KE increases