# Help with Change in Electric Potential Ranking Task. I am looking for some conceptual help with this physics problem.?

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I am looking for help in understanding the why:

The problem asks to rank the following electrical potentials, “In the diagram below, there are two charges of +q and −q and six points (a through f) at various distances from the two charges.(Figure 1) You will be asked to rank changes in the electric potential along paths between pairs of points”. Here is the Figure:

http://session.masteringphysics.com/problemAsset/1…

I am not simply looking for an answer. I know the answer is c to b > d to a > c to d > f to e = b to a > c to e. I am looking for an explanation into solving problems like this and some conceptual help. I understand that as you get closer to the + source charge the potential increases and as you get closer to the – source charge it decreases. I also understand that equidistant point charges = 0V. I am having trouble with the rest and not quite getting it. Could someone maybe use the equation V = (kq_1)/r_1 + (kq_2)/r_2 and input numbers to help this make more sense?

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All help is greatly appreciated! I am happy to give a good answer that helps explain this a best answer!

• Well, you should know that

potential = algebraic Σ k*q/d

so at c and d, which are both equidistant from the two charges,

the potential must be zero.

a and b, closer to the positive charge than to the negative,

must have positive potential (the negative charge’s potential

is diluted due to its greater distance)

and similarly, e and f, closer to the negative charge than

to the positive charge, must have negative potential.

The only remaining question, then, is:

Does a or b have greater potential?

If each square is 1 unit of distance, then

Vb = k*q*(1/1 – 1/5) = 4kq/5

and

Va = k*q*(1/2 – 1/6) = kq/3

so the ranking of points from highest potential to lowest must be

b – a – cd – f – e

+ ……. 0 …… –

I’m guessing you weren’t asked to rank ALL POSSIBLE paths, but just the specific ones you list — cb, da, cd, fe, ba, ce

Well, I’d say

Vcb = Vb – Vc = 4kq/5 – 0 = 4kq/5

Vda = Va – Vd = kq/3 – 0 = kq/3

Vcd = Vd – Vc = 0 – 0 = 0

Vfe = Ve – Vf = -4kq/5 – -kq/3 = -7kq/15

Vba = Va – Vb = kq/3 – 4kq/5 = -7kq/15

Vce = Ve – Vc = -4kq/5 – 0 = -4kq/5

has them ranked (since I used your order). 