A charge of 28.0 nC is placed in a uniform electric field that is directed verticall upward and that has a magnitude of 4.00*10^4 N/C. What work is done by the electric force when the charge moves (a)0.450 m to the right (b) 0.670 m upward (c) 2.60 m at an angle of 45 degrees downward from the horizontal?

Question

A charge of 28.0 nC is placed in a uniform electric field that is directed verticall upward and that has a magnitude of 4.00*10^4 N/C.  What work is done by the electric force when the charge moves (a)0.450 m to the right (b) 0.670 m upward (c) 2.60 m at an angle of 45 degrees downward from the horizontal?

I am totally lost on this question! help please!

drwls · Accepted Answer

a) zero, since movement is perpendicular to the force

b) Work = (force) x (distance)
= E Q x 0.67 m
E is the field strength and Q is the charge
c) - E Q x(2.60 m sin 45)
The sin 45 factor is for the component of the motion along in the direction of the force. The minus sign is there because work is done against the field.