How Does a Parallel Plate Capacitor with Different Dielectrics Calculate Keq?

[Kathi201]

A parallel plate capacitor is constructed such that there are two different dielectric materials in between the plates, each occupying half the space in between . Still this capacitor can be modeled as a simple parallel plate capacitor with same total area A, plate separation d, but with an equivalent dielectric constant Keq. What is Keq? (Your final answer will be in terms of the two K's, and for a start, you should think in terms of potential being same all over the surface for any given plate)

_________
l l l
l K1 l K2 l d
l____l____l
with an area A (it is supposed to look like a square with two sides K1 and K2)

I know the capacitance of a parallel-plate capacitor filled with a dielectric is C = KEoA / d
But I am not quite sure what to do if K is split into K1 and K2 as displayed in the picture above. Do I set them equal to each other?

Any help would be appreciated!

 

[tiny-tim]

cheeseburger?

A parallel plate capacitor is constructed such that there are two different dielectric materials in between the plates, each occupying half the space in between …

Hi Kathi!

Which half?

(I can't understand your diagram)

Is this like a cheeseburger, or like two different ordinary burgers side-by-side?

 

[baba89]

To calculate the equivalent dielectric constant (Keq) for this parallel plate capacitor, we can use the concept of electric potential. The potential difference between the plates is determined by the electric field and the distance between the plates. In this case, the electric field is the same for both dielectric materials, since they are in parallel, but the distance between the plates is different for each material.

We can use the equation for electric potential, V = Ed, to calculate the potential difference for each dielectric material. Since the potential is the same for both materials, we can set these two equations equal to each other:

V1 = E1d1
V2 = E2d2

Since we know that the electric field is the same for both materials, we can divide both equations by E and rearrange to solve for d:

d1 = V1 / E
d2 = V2 / E

Now, we can substitute these values into the equation for capacitance:

C = KEoA / d

For the first dielectric material, we have:

C1 = K1EoA / (V1/E)

And for the second dielectric material, we have:

C2 = K2EoA / (V2/E)

Since the capacitance is the same for both materials, we can set these two equations equal to each other and solve for the equivalent dielectric constant (Keq):

C1 = C2
K1EoA / (V1/E) = K2EoA / (V2/E)
K1 / V1 = K2 / V2
K1 / K2 = V1 / V2

Therefore, the equivalent dielectric constant for this parallel plate capacitor is:

Keq = K1 / K2

This means that the equivalent dielectric constant is the ratio of the dielectric constants of the two materials. In this case, it is the same as the ratio of the distances between the plates for each material.

I hope this helps to answer your question. Let me know if you have any further questions or need clarification.