A moving particle encounters an external electric field that decreases its kinet

Question

A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particle moves from postition A to position B. The electric potential at A is -55.0V, and the electric potential at B is +27.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.

Explanation / Answer

Energy conservation => loss in kinetic energy = gain inelectrostatic potential energy change in KE = 9520eV - 7060eV = 2460eV U = qV => change in U = q x (change in electric potential) 2460eV = 2460 x 1.6 x 10-19 J = q *(27-(-55)) => q = + 4.8x10-18 C

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