Consider the arrangement of two fixed point charges, equal in magnitude

[Physics1988]

Consider the arrangement of two fixed point charges, equal in magnitude...

Consider the arrangement of two fixed point charges, equal in magnitude, shown in the figure.
Which of the following statements are correct for the initial motion of a third charge if it is released from rest in the vicinity of the two charges shown?
IMAGE: http://i51.tinypic.com/2akeovn.gif

A negative charge at point d will accelerate down. (false)
A negative charge at point e will accelerate up. (true)
A negative charge at point b will accelerate down.(true)
A positive charge at point a will accelerate toward the lower-left. (false)
A negative charge at point c will accelerate toward the lower-right. (true)

I have no idea why this is wrong. I thought it would be an easy, 30 second question but I just can't seem to get it. The answers to the left is what I think is right, but is apparently wrong.

PLEASE, an explanation!

Ok so this is my thought process. A negative charge at D will be repulsed by the negatively charged point. A negative charge at e will be attracted to the positively charged point. B will be both attracted to the positive point and repulsed by the negative. If I draw vectors, the vector sum of A is on the top left and for C, it is on the lower right.Just realized that this might be in the wrong section. Sorry, first time posting. Please feel free to move.

 

[Delphi51]

Welcome to PF, 1988.

I would say the answers are correct except for the last question, the negative charge at c. If you draw the force vectors on c, one away from the negative charge and the other toward the positive charge, you will see that the horizontal parts cancel out and the total force is straight down.

 

[Physics1988]

Thank you for the welcome!

And yes you are correct, that was the one that was wrong.

I thought they meant that it ends up somewhere in the lower-right, which is why I said it was true. Turns out they are asking if the DIRECTION is lower-right, which it is not.

I also found this :
http://tinypic.com/r/fxs4yo/7

So I guess any point would have to be in the positive-to-negative trajectory?

 

[Delphi51]

Yes, the direction is the thing. The fxs4yo/7 link is to an electric field line diagram, not the same thing as a vector direction. The direction of the electric field at any point would be tangential to the field line at that point.

So I guess any point would have to be in the positive-to-negative trajectory?

I don't follow this terminology . . . trajectory means "flight path" to me. Here the velocity vector would be tangential to the trajectory. I can't quite picture how the acceleration vector would be related.