Electric Potenital engery trouble!

Hello everyone i have the following picutre:
http://www.webassign.net/hrw/24_02.gif
And i'm having troubles with some of the questions I answered some of them and i was wondering if u could check to see if i did those right and to see if you can help me on the one i didn't do:

Figure 24-22 shows three paths along which we can move positively charged sphere A closer to positively charged sphere B, which is held fixed in place.

Well the rule is for part (a) Electric Potential alwasy decreases in the direction of the eletric field. But we have 2 positved charged particles. How is there an eletric field? does an E-field go from positive to negative? So what direction would the e-field even be pointing?
Figure 24-22
(a) Would sphere A be moved to a higher or lower electric potential?
higher
lower
(b) Is the work done by our force positive, negative, or zero?
positive
[negative]
zero
(c) Is the work done by the electric field due to B positive, negative, or zero?
positive
negative
zero
(d) Rank the paths according to the work our force does, greatest first (use only the symbols > or =, for example 1=2>3).
1 = 2 = 3

Thanks! I'm also confused on (c), first i need a way to figure out what direction the e-field is going then i'll be okay maybe!

 

thanks!! so since sphere A is moving towards its electric field, its Electric potenial must be decreasing also because B is fixed, does that mean work is zero? because u need movement for work to happen right?

 

Thahnks for the explanation! it helped alot, but i'm still confused on somthing...
I understand part (a), I also understand part (d). But The answers for part (b) and (c) i don't get. The answer for (b) Is the work done by our force positive, negative, or zero? Answer: Positive.
Okay, "Our" force, i'm assuming is the electric force on A by B. So here is my diagram.
F1 = Force of moving the charge.
F2 = Electric Feild from B on A.
d = distance A is moving.

<---F1--(A)--F2--->
<---d---
W = FD
W = qEdcos(x)
W = (+1.6C)(-1N/C)(+1m)(cos(180));
W = (+)(-)(+)(-) = +
ohhh, i think i just explained it by asking the question, did I prove that right?

Okay for (c)
Is the work done by the Electric field due to B positive, neg, or 0.
Well isn't this the same as part (b)?
The efield on A is pushing it away, A is moving in the opposite direction of the E-field, so wouldn't it also be:
W = (+)(-)(+)(-) = + ?

THanks again!