# Calculating the Electric field in the capacitor

I am trying to understand capacitors and have come across the below example. What I would like to understand is how to calculate the electric field at some distance $x$ within the capacitor.

The image is of two circular disks as the capacitor plates with radius $R$. These are attached to a dielectric with a dielectric constant of $k_1$ separated by another dielectric with constant $k_2$. The plates are attached to a voltage source with voltage $V$.

I am reading through this link explaining capacitors and dielectrics with a similar example at problem 27.19, however with only one dielectric. I am getting a bit stuck on if it is correct to use the following equation, and if so, what is the integral result (I'm not very good at integration).

$$\Delta V = V_3 - V_2 - V_1 = - \int_{plate\,1}^{plate\,2} E \,\,\delta y = - \int_{0}^{a} \frac{E_o}{k_1} - \int_{a}^{b} \frac{E_o}{k_2} -\int_{b}^{c} \frac{E_o}{k_1}$$

where $V_3$ is the voltage in region $c$, $V_2$ is the voltage in region $b$, and $V_1$ is the voltage in region $a$. $E_o$ is the electric field without a dielectric and equal to $\frac{\sigma}{\epsilon_o}$ and $\sigma = \frac{Q_{total}}{{A_{total}}}$ where:

$$Q_{total} = \frac{{\epsilon_o}{A}}{d} \Delta V$$

Do I need to consider that the electric field at some point $x$ from a circular disk is given as:

$$E = \frac{{\sigma}}{2\epsilon_o}\biggl(1 -\frac{x}{\sqrt{x^2+R^2}}\biggr)$$

Could someone advise if this is the correct method or if I am missing something, and point me in the direction of how to solve the integral. If the disk calculation is more complex, how would the electric field be calculated if they were parallel capacitor plates.

Thanks.

So your equation for the electric field $E = \dfrac{{\sigma}}{2\epsilon_o}\biggl(1 -\frac{x}{\sqrt{x^2+R^2}}\biggr)$ tends to $E = \dfrac{{\sigma}}{2\epsilon_o}$ as the assumption is that $x\ll R$.
• Thanks. So I calculate the integral with $E_0 = \frac{\sigma}{2\epsilon_0}$? How do I go about calculating the electric field if $x = R$ – ED9909 Jan 6 '18 at 11:20