Problem 11-71 Grav & Elec Potential - Part 4 - C

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.

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

Diagram of proton moving toward planet.

No. The gravitational force is doing work against the electric field. (In the absence of gravity the charge would move the other way.) Does the electrical PE increase or decrease?

Choose one.