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Posted by on Tuesday, February 16, 2010 at 6:16pm.

PLEASE EXPLAIN

A charge moves a distance of 1.7 cm in the
direction of a uniform electric field having
a magnitude of 200 N/C. The electrical
potential energy of the charge decreases by
9.69095 × 10−19 J as it moves.
Find the magnitude of the charge on the
moving particle. (Hint: The electrical poten-
tial energy depends on the distance moved in
the direction of the field.)
Answer in units of C.

  • physics - , Tuesday, February 16, 2010 at 6:29pm

    work done = force * distance = change in energy

    Work = Q E d
    9.69095*10^-19 = Q (200) (1.7*10^-2)

    Q is positive because its potential energy decreases as it moves with the field (the field pushes it down the potential).

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