I am trying to understand capacitors and have come across the below example.enter image description hereWhat 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.


The capacitor being considered, although you would guess that from the diagram) is one for which the linear dimension of a plate is much larger than the separation of the plates and so the the electric field between the plates is assume to be uniform.

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 $x\ll R$.

If this approximation is not made the calculation of capacitance is much more complicated as you do have to consider the fact that the electric field escapes from the confines of the body of the capacitor.

  • $\begingroup$ Thanks. So I calculate the integral with $E_0 = \frac{\sigma}{2\epsilon_0}$? How do I go about calculating the electric field if $x = R$ $\endgroup$ – ED9909 Jan 6 '18 at 11:20

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