平行板的电势

Suppose that we have two infinitely large parallel plates, one with a positive charge and one with a negative charge. Suppose that both have the same charge density $\sigma$. Let the two plates be separated by a distance $d$.

The electric field due to the positively charged plate is: $$\vec{E}=\frac{\sigma}{2\epsilon_0}\vec{r}$$ where $\vec{r}$ is the unit vector pointing away from the positive plate.

The electric field due to the negatively charged plate is: $$\vec{E}=\frac{\sigma}{2\epsilon_0}\vec{s}$$ where $\vec{s}$ is the unit vector pointing toward the negative plate.

When we are located between the two plates, the vector pointing toward the negative plate $(\vec{s})$ and the vector pointing away from the positive plate $(\vec{r})$ are pointing in the same direction! Thus, the electric fields on the two plates build on each other, so the total electric field is: $$\vec{E}=\frac{\sigma}{\epsilon_0}\vec{r}$$ where $\vec{r}$ is the unit vector pointing from the positive plate to the negative plate.

However, when we are located beyond the plates, such that both plates are on the same side of the charge, then the vectors $\vec{r}$ and $\vec{s}$ point in opposite directions and thus cancel out. Therefore, we have: $$\vec{E}=\vec{0}$$ Now, define the $x$-coordinate of a point $P$ in space to be the directed distance from the negative plate to the point $P$. The direction that goes toward the positive plate will be regarded as positive. For example, any point on the positive plate will have an $x$-coordinate of $d$.

For convenience, set the potential at $x=0$ to be $V=0$. Of course, since we only care about potential differences, this is always allowed.

Now, the entire space beyond the negative plate has no electric field, meaning that the whole space is an equipotential. But what is this potential? It must be the same as that of the negative plate, which is $0$! Thus, for $x\lt 0$, we have $V=0$.

Now, consider a value of $x$ such that $0\le x\le d$. Now, we use an integral to calculate the voltage: $$V=\int_{0}^{x}\vec{E}\cdot\vec{dl}$$ Now, we may choose the most convenient path: a straight segment perpendicular to the negative plate. Thus, we may rid ourselves of the vector notation: $$V=\int_{0}^{x}E dx$$ Since $E$ is constant, regardless of $x$, we may compute this very easily: $$V=\int_{0}^{x}E dx$$ $$V=Ex$$ $$V=\frac{\sigma x}{\epsilon_0}$$ Thus, for $0\le x\le d$, we have $V=\frac{\sigma x}{\epsilon_0}$.

Last, we consider the space beyond the positive plate. This is an equipotential, since the electric field is zero everywhere. Thus, the potential here must be the same as that of the positive plate. This potential may be found by plugging $x=d$ into the expression above. Thus, for $x\gt d$, we have $V=\frac{\sigma d}{\epsilon_0}$.

Therefore, we have calculated the electric potential due to a pair of charged plates at any point in space. Here is the summary: $$V=0 (x\lt 0)$$ $$V=\frac{\sigma x}{\epsilon_0} (0\le x\le d)$$ $$V=\frac{\sigma d}{\epsilon_0} (x\gt d)$$