Trouble with negative sign in this Potential Difference problem

[guyvsdcsniper]

Homework Statement

Let A=(x1,y1) and B=(x2,y2) be two points near and on the same side of a positively charged sheet with a uniform surface charge density. The electric field E⃗ due to such a charged sheet has the same magnitude everywhere and points away from the sheet, as shown in

Relevant Equations

V= -E*ds

So I know that E = -ΔV/Δs. If I wanted to solve for change in potential I could rearrange this equation and get Δ = -E*ds. With that information I believe I can solve the problem below. But in both solutions provided below, the negative sign goes away. Now I know I can pull the E out because it is uniform. Mylab does not go into detail on why the negative goes away. The hand written work includes the negative up until the 2nd to last step and I don't see how they got rid of it.

Any help with why the negative goes away?

 

[BvU]

Hi,

You have a problem when you use 1,2 and A,B all together. Pick one of these two sets and use it consistently.

 

[Steve4Physics]

So I know that E = -ΔV/Δs

Some thoughts…

You are asked to find $V_{AB}$ which is defined as $V_A – V_B$. We can interpret this as the potential of A relative to B. (In the diagram shown, we expect this to be a positive value because point A is nearer to the positively charged sheet than point B.)

But (I think) your use of $ΔV$ is such that $ΔV$ means $V_B – V_A$. This means ΔV is the change in potential moving from A to B; it is the potential of B relative to A. So $ΔV = -V_{AB}$.

And I agree with @BvU. mixing A, B, 1 and 2 seems to be a potential (pun intended) source of confusion.

 

[haruspex]

This line is wrong:
$\vec{dl}=(x_2-x_1)\hat x+(y_2-y_1)\hat y$, where $A=(x_1,y_1)$ etc.
On the left, $\vec {dl}$ is a small element of the path, while on the right,
$(x_2-x_1)\hat x+(y_2-y_1)\hat y$ is the whole path.
It should read $\vec{dl}=dx\hat x+dy\hat y$.
$\vec E.\vec{dl}=dy|E|$
$V_{AB}=-\int_B^A|E|dy=\int_A^B|E|dy=(y_2-y_1)|E|$

This confusion in the text in the image effectively led to the bounds being ignored in the integration and thereby treated as $_A^B$.