Multiple dielectrics in one capacitor

[Rake-MC]

Hi guys, I'm absolutely, completely and utterly stumped with this one question I've been given..

Here is the exact question:

Given that you possesses the following materials:
(i) two sheets of copper of uniform thickness;
(ii) one sheet of mica of uniform thickness 0.10 mm and dielectric constant κ= 6;
(iii) one sheet of glass of uniform thickness 2.0 mm and dielectric constant κ = 7;
(iv) one sheet of paraffin wax of uniform thickness 1.0 cm and dielectric constant
κ = 2;
determine which sheet, or combination of sheets, will produce a parallel plate capacitor of
the largest capacitance when placed between the copper sheets. Assume that all sheets
have the same shape.

Relevant equations that I have are:
C=(kappa)(epsilon)(Area)/(Distance)
V=V[initial]/kappa
but there are probably plenty more. I have equations for potential energy and such, but I'm not sure if they are necessary for the question.

attempt at a solution

Well as I said, I am completely stumped. I can calculate the capacitance of each dielectric on its own in the capacitor, but the combination ones are the ones I'm stuck with. This is what I've got so far {probably way off track}.

C[mica] = 6(epsilon)(area)/0.1x10^-3
C[glass] = 7(epsilon)(area)/2x10^-3
C[wax] = 2(epsilon)(area)/0.01

Can I ignore epsilon and area because they are common in all the equations?

for the dielectrics next to each other;

C[effective] = C[no dielectric]*k[1]*k[2]

This is most likely wrong. I know that for one dielectric, you can simply say that C[effective] = C[no dielectric]*k
where k = kappa

But I simply don't know how to derive a formula for multiple dielectrics.

Also, I do have knowledge of the theory of how dielectrics work, ie. how they generate opposing electric fields to that of the capacitor, but I am still 100% stuck with this question..Thanks in advance

 

[Irid]

If you have multiple sheets of dielectric stacked on each other, you can imagine that each sheet is a distinct capacitor and these capacitors are connected in series. The total capacitance C is then

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}$$

 

[baba89]

for any help or guidance you can provide.

Dear student,

Thank you for reaching out for help with this question. The concept of multiple dielectrics in one capacitor can be a bit confusing, but I will try my best to explain it to you.

First, let's review the relevant equations for capacitance:

C = Q/V = εA/d

Where:
C = capacitance (in Farads)
Q = charge (in Coulombs)
V = voltage (in Volts)
ε = permittivity of the material (in Farads/meter)
A = area of the plates (in square meters)
d = distance between the plates (in meters)

Now, let's address your attempt at a solution. You are correct in calculating the capacitance for each individual dielectric using the equation C = κεA/d. However, when it comes to multiple dielectrics, you cannot simply add the capacitances together. This is because the electric fields generated by each dielectric will interact with each other, affecting the overall capacitance.

To solve this problem, we need to use the concept of equivalent capacitance. This means that we can replace multiple capacitors with one equivalent capacitor that has the same overall capacitance. In this case, we can think of the copper sheets as the plates of a capacitor, and the dielectric sheets as the dielectric material between the plates.

To calculate the equivalent capacitance, we can use the formula:

1/C[equivalent] = 1/C[1] + 1/C[2] + 1/C[3] + ...

Where C[1], C[2], C[3], etc. are the capacitances of each individual dielectric.

Using this formula, we can calculate the equivalent capacitance for the given materials:

1/C[equivalent] = 1/C[copper] + 1/C[mica] + 1/C[glass] + 1/C[wax]

= 1/ε[copper]A/d + 1/κ[mica]ε[mica]A/d + 1/κ[glass]ε[glass]A/d + 1/κ[wax]ε[wax]A/d

= 1/(ε[copper] + κ[mica]ε[mica] + κ[glass]ε[glass] + κ[wax]ε[wax]) A/d

= 1/ε[equivalent] A/d