Problem 11-71 Grav & Elec Potential - Part 5 - 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
\(F_E = qE \quad F_g = mg \\ F_g = mg = (1.673 \times 10^{-27} )(1.3 \times 10^5) = 2.175 \times 10^{-22} \; N \\ F_E = qE = (1.602 \times10^{-19} )(2.47 \times 10^{-4}) = 3.957 \times 10^{-23} \; N \)
\(E_{K1} +E_{P1} +U_1 = E_{K2} +E_{P2} +U_2 \\ E_{K1} = 0\)
Correct.
Therefore
\(E_{K2} = (E_{P1} - E_{P2}) + (U_1 - U_2)\)
Before continuing solve for \(E_{K2}\) in terms of the change in the potentials.