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Posted by on Wednesday, July 4, 2012 at 4:04pm.

A moving particle encounters an external electric field that decreases its kinetic energy from 9100 eV to 6630 eV as the particle moves from position A to position B. The electric potential at A is -59.5 V, and the electric potential at B is +26.1 V. Determine the charge of the particle. Include the algebraic sign (+ or −) with your answer.

  • physics - , Wednesday, July 4, 2012 at 4:22pm

    ΔKE=-ΔPE= q•Δφ.
    9100-6630=2470 eV=2470•1.6 •10^-19 =3.95 •10^-16 J.
    |Δφ|=|φ1- φ2|=|-59.5-26.1|=|-85.6|=85.6 V
    q=3.95 •10^-16/85.6 = + 4.62•10^-18 C

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