A moving particle encounters an external electric field that decreases its kinetic energy from 9160 eV to 7160 eV as the particle moves from position A to position B. The electric potential at A is -58.0 V, and that at B is +22.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.

Answer 1

q*deltaV=deltaKE
take the change in KE (in ev), convert to Joules, then solve for q
deltaV = positive eighty v
deltaKE=final-initial (it will be negative, it lost KE), indicating it slowed, ie positive charge