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Homework Statement
A particle with a charge of + 4.20 nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is found to be +1.50 * 10^-6 J
What work was done by the electric force?
Homework Equations
W= -U = qEd
The Attempt at a Solution
I know the magnitude of the answer is 1.5 * 10^-6, but I'm having trouble with the sign. Since the field would naturally move the particle to the left, shouldn't the work be done by the electric force be negative? The force is not moving the particle against the direction of motion so it wouldn't be positive.