Electric field in a capacitor with multiple dielectrics

[Bhope69199]

Hi,

I am trying to understand capacitors and have come across the example in the attached image.

What I would like to understand is how to calculate the electric field at some distance $x$ within the capacitor. With $x>>R$ , $x<R$ and $x=R$ .

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 http://teacher.nsrl.rochester.edu/phy122/Lecture_Notes/Chapter27/chapter27.html 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).

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

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:

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

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 with $x>>R$ , $x<R$ and $x=R$ . Where $R$ is now half the total length of the parallel plates.

Thanks.

 

[Baluncore]

Since the changes in dielectric constant are parallel and so equipotentials, you can model it as three capacitors Ca, Cb and Cc in series. Are you expected to model the fringing capacitance external to the dielectric?

Since the same charge is applied to all capacitors in series v = q/C, the voltage will be shared, inversely to capacitance. Express the voltage gradient across each and construct a piecewise function.

 

[Delta2]

The integrals are easy if you consider $E_0$ to be constant inside each dielectric. First integral equals $\frac{E_0}{k_1}(a-0)$ second $\frac{E_0}{k_2}(b-a)$ and third $\frac{E_0}{k_3}(c-b)$

But the above case is the case where $d<<2R$, cause that's the case where you can consider $E_0$ to be almost constant (as approximation). If $d\geq 2R$ then that approximation about $E_0$ is not as good, and the larger the d becomes, the less accurate is that approximation.

Btw your equation about $Q_{total}$ doesn't seem to be dimensionally correct. check that equation what was it you were really trying to write there?