physics
posted by Jonathan .
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.

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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