A movig particle encounters an external electric field that decreases its kineti

Question

A movig particle encounters an external electric field that decreases its kinetic energy from 9960 eV to 7700 eV as the particle moves from position A to position B. The electric potential at A is -53.0 V, and that at B is +34.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.

Explanation / Answer

A movig particle encounters an external electric field that decreases its kinetic energy from 9400 eV to 8100 eV as the particle moves from position A to position B. The electric potential at A is -51.0 V, and that at B is +36.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer. ?Energy = Q?V ==> Q = ?Energy/?V = 1300/87 = 14.9425 elementary charges or 14.9425*1.6022E-19 = 2.39409E-18 C. The sign of Q is positive because KE decreases and thus PE increases with increasingly positive voltage.

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