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Your answer is partially correct. Try again. A moving particle encounters an external electric field that decreases its kinetic energy from 9100 eV to 6990 eV as the particle moves from position A to position B. The electric potential at A is-41.0 V, and that at B is +23.0 V. Determine the charge of the particle. Include the algebraic sign (+ or) with your answer Higher potential Lower potential 8 Number +5.48E-18 UnitsCI don't know why the answer is wrong, I got K(final) = 1.1072x10^-15 J and K(initial) = 1.4576x10^-15J, with the work energy theorm I got q=(Kf-Ki)/(Vi-Vf) = +5.48x10^-18 C

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16 IS ORnRi ki-nttäl kinehc enas? IS -J8

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