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 9160 eV to 6010 eV as the particle moves from position A to position B. The electric potential at A is -65.0 V, and that at B is +40.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.

A moving particle encounters an external electric field that decreases its kinetic energy from 9160 eV to 6010 eV as the particle moves from position A to position B. The electric potential at A is -65.0 V, and that at B is +40.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.

Explanation / Answer

dKE + dPE = 0

dPe = q dV

(6010-9160) + q*(40+65) = 0

q=30 e = 30*1.6E-19 = 4.8E-18 C

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