Problem with line integrals for electric potential

[WWCY]

Homework Statement

I have a problem understanding the equation
$$\Delta V = -\int_{a}^{b} \vec{E} \cdot d \vec{l}$$

In the case of a parallel plate capacitor whereby the positive plate is placed at $z=t$ while the negative is at $z = 0$, my integral looks like
$$\Delta V = -\int_{0}^{t} E(-\hat{z}) \cdot dz({\hat{z}})$$
since E points down, but my path integral goes up.
However, if I switch the position of my plates, the integral looks like
$$\Delta V = -\int_{t}^{0} E(\hat{z}) \cdot [-dz({\hat{z}})]$$
since field points up and my line integral goes down.

Essentially, my integrands look the same but the limits are flipped, which means that the signs are different. But doesn't the original equation mean something along the lines of "the potential at b with respect to a", which means that changing the coordinates of the positive and negative plates shouldn't change the answer?

Some clarification is greatly appreciated!

Homework Equations

The Attempt at a Solution

 

[haruspex]

which means that changing the coordinates of the positive and negative plates shouldn't change the answer?

I followed it all until that step.
Perhaps you need to be careful with what you mean by "the potential at a" etc. Are you thinking of that as a point in space or the position of "plate A"? Are you being inconsistent about that?

 

[WWCY]

Thank you for the response

I followed it all until that step.
Perhaps you need to be careful with what you mean by "the potential at a" etc. Are you thinking of that as a point in space or the position of "plate A"? Are you being inconsistent about that?

Apologies, let me try to rephrase the whole thing a bit. What I meant was "potential of a capacitor plate at point b, with respect to the potential of another capacitor plate at point a". So for example, I could have a plate at $z = 3$ lying parallel to the $xy$ plane, and another plate at $z=0$ that lies on the $xy$ plane.

Say the plate on top is with positive charge. My potential difference integral would look like
$$\Delta V = -\int_{0}^{3} E (-\hat{z}) \cdot dz(\hat{z})$$
since field points in $-\hat{z}$ and the integral path is in $\hat{z}$ from the (-) plate to the (+) plate. But if I decide that the bottom plate should be positive, and the top negative,
$$\Delta V = -\int_{3}^{0} E (\hat{z}) \cdot dz(-\hat{z})$$
which gave me the sign error. I would expect potential of the (+) plate to always be positive w.r.t the (-) one. What am I getting wrong here?

Thank you for your assistance!

 

[Dr Dr news]

By definition the direction of the electric field is the direction of the force on a positive test charge so in a two-plate capacitor, the electric field is always directed from the positive plate to the negative plate. If you are consistent in the algebraic sign of your coordinate axis, the difference in potential should be consistent.

 

[haruspex]

which gave me the sign error

ok, I see the problem; it's the $dz(-\hat z)$. There is no basis for the minus sign.
if you switch the sign on dz as well as switching the bounds then nothing changes. If you think of the integral as a sum, all you have done is reversed the order of summation. This done not change the sign of the result. So for the reversed limits to have the desired effect you must leave dz as positive.

 

[WWCY]

Thank you both for the replies

ok, I see the problem; it's the $dz(-\hat z)$. There is no basis for the minus sign.
if you switch the sign on dz as well as switching the bounds then nothing changes. If you think of the integral as a sum, all you have done is reversed the order of summation. This done not change the sign of the result. So for the reversed limits to have the desired effect you must leave dz as positive.

Is this to say that the direction of path is determined by integral limits, while the "axis" along which I integrate is defined by $d\vec{l}$? Also, am I right to assume that this applies to line integrals in general?

Thank you.

 

[haruspex]

am I right to assume that this applies to line integrals in general?

It applies to integrals in general.
$\int_a^bf(x).dx=-\int_b^af(x).dx=\int_b^af(x).(-dx)$
So reversing the bounds and switching the sign on the dx has no net effect, but you intended a different result - namely, P_a-P_b instead of P_b-P_a.