Problem 11-71 Grav & Elec Potential - Part 6

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.

\(E_{K1} +E_{P1} +U_1 = E_{K2} +E_{P2} +U_2 \\ E_{K1} = 0 \\E_{K2} = (E_{P1} - E_{P2}) + (U_1 - U_2)\)

You should have
\(E_{K2} =  -(E_{P2}- E_{P1} ) - (U_2 - U_1)\)

In this problem we have {BE CAREFUL THIS IS TRICKY}

(A)   \((E_{P2}- E_{P1} )\) positive and \((U_2 - U_1)\) negative

(B)   \((E_{P2}- E_{P1} )\) negative and \((U_2 - U_1)\) positive

(C)   \((E_{P2}- E_{P1} )\) positive and \((U_2 - U_1)\) positive

(D)   \((E_{P2}- E_{P1} )\) negative and \((U_2 - U_1)\) negative