184_notes:e_parallel_plates

We can find the electric field between the plates using Gauss's Law (just like you did in Project 5). We know that for two parallel plates, there is an electric field in the middle that points directly from the positive plate to the negative plate (except near the edges where the field bends slightly out). Outside of the plates, the electric field is zero because the contributions from the negative and positive plates will cancel.

If we draw a cylinder around the top plate, we can find an expression for the electric field using Gauss's Law. So starting with: $$\int \vec{E} \cdot \vec{dA}=\frac{Q_{enc}}{\epsilon_0}$$ We will assume that the plates are large and our Gaussian surface is in the middle of the plate. This means that the electric field will be uniform and point directly downward. The $\vec{dA}$ for the cylinder would point straight down on the bottom surface, straight up on the top surface, and radially outward on the side.

$$\int \vec{E_{bot}} \cdot \vec{dA_{bot}}+\int \vec{E_{top}} \cdot \vec{dA_{top}}+\int \vec{E_{side}} \cdot \vec{dA_{side}}=\frac{Q_{enc}}{\epsilon_0}$$

Because the electric field is perpendicular to the dA vector on the side of the cylinder, that part of the electric flux is zero. Since the electric field above the top plate is zero, the electric flux through the top surface is also zero. This leaves only the bottom surface in the equation. $$\int \vec{E_{bot}} \cdot \vec{dA_{bot}}=\frac{Q_{enc}}{\epsilon_0}$$ For the bottom surface, $\vec{E}$ and $\vec{dA}$ are parallel, so the dot product turns into multiplication. $$\int E_{bot}*dA_{bot}=\frac{Q_{enc}}{\epsilon_0}$$ Then the electric field is constant everywhere along the bottom area, so we can pull it out of the integral, which leaves: $$E_{bot} \int dA_{bot}=\frac{Q_{enc}}{\epsilon_0}$$ $$E_{bot} A_{bot}=\frac{Q_{enc}}{\epsilon_0}$$ Finally, we can write the charge enclosed in terms of the charge density ($\sigma$) on the plate and the area enclosed by the cylinder ($A$). $$E_{bot} A_{bot}=\frac{\sigma A}{\epsilon_0}$$ Because the cylinder is the same size at all points, $A_{bot}=A$, which leaves: $$E_{bot}=\frac{\sigma}{\epsilon_0}$$ So the electric field in the middle of the capacitor has a magnitude of $E=\frac{\sigma}{\epsilon_0}$

  • 184_notes/e_parallel_plates.txt
  • Last modified: 2017/08/10 10:49
  • by dmcpadden