Question

A charge of -4.00x10^-6 C is placed at a point in space where the electric field is directed toward the right and has a magnitude of 8.00 104 N/C. What are the magnitude and direction of the electrostatic force on this charge?

Answer 1

E = F/q

start with that for the magnitude question.

Answer 2

in terms of direction, you draw field diagrams with field lines moving away from a +ve charge, as that is how a +ve charge would move in such a field. in this case, it seems that the field lines would move left to right, but would be equally spaced. a +ve charge would also go in that L-R direction with no movement in the vertical direction as the field is uniform in that direction.

it is very much worth drawing this out, as stupid as that might sound, because all the brilliant stuff in electromagnetism starts with Maxwells equations and Gauss' law and so on -- and simple stuff like field diagrams are incredibly useful in that regard.

Answer 3

I've tried dividing them by each other and I am getting an incorrect answer. What am I doing wrong?

Answer 4

might the problem be with 8.00 104 N/C? are you using that as 8 x 10 ^ (4) ? ie 80 000N/C

Answer 5

Yes, that is what I am using. Is that not right?

Answer 6

So you're getting -0.32N, right?

Answer 7

Yes. It says it's incorrect

Answer 8

Apparently the answer is +.32N. Why is the sign changed?

Answer 9

"magnitude". the magnitude of the force is .32.

the sign indicates the direction the force will act in.

is that any help?

Answer 10

Sort of. So when it asks me whether the direction will be up, down, left or right, how do I decide. I know the sign will tell me but I am still kind of confused

Answer 11

field points to right. that means a +ve charge placed in that field will move to the right if it is free to move. but in this case, your charge is -ve.