- #1
WWCY
- 479
- 12
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!