PLEASE EXPLAIN

Question

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.

Answer 1

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).