Capacitors with partial dielectrics

In summary: I need somebody who knows what they're doing to help me out!In summary, the capacitance of the capacitor with the dielectric is C(f) as a function of f.
  • #1
clope023
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[SOLVED] Capacitors with partial dielectrics

Homework Statement



Consider two parallel-plate capacitors identical in shape, one aligned so that the plates are horizontal, and the other with the plates vertical

horizontal:

14985_a.jpg


vertical:

14985_b.jpg


The horizontal capacitor is filled halfway with a material that has dielectric constant . What fraction of the area of the vertical capacitor should be filled (as shown in the figure) with the same dielectric so that the two capacitors have equal capacitance? Express your answer in terms of K


Homework Equations



e = epsilon_0

Ch (horizontal capacitance) = 2KAe/d(K+1)

Cv (vertical capacitance) = Ae(1-f+Kf)/d

The Attempt at a Solution



I've already found the capacitance of both capacitors, however it's asking for a fraction of f by itself, I've tried a few combinations, a ratio, an equivalence and I can't seem to cancel out enough variable to get f by itself in one side such that I have an answer in terms of K, if anybody could point me in the right direction, I would greatly appreciate it.
 
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  • #2
Hi clope023,

They want Ch=Cv, so you can set the right hand sides equal to each other:

2KAe/d(K+1) = Ae(1-f+Kf)/d

and solve for f. When you solve for f, all of the other unknowns will cancel except for k.
 
  • #3
alphysicist said:
Hi clope023,

They want Ch=Cv, so you can set the right hand sides equal to each other:

2KAe/d(K+1) = Ae(1-f+Kf)/d

and solve for f. When you solve for f, all of the other unknowns will cancel except for k.

hey thanks! got the answer, now I actually have another dielectric problem, I've more or less solved it utlizing the meaning of C, but they want the answer in terms of C as a function of f, let me show you:

39460.jpg


I know the total Capacitance = Cdieletric and Cempty

Cdielectric = KeAf/d

Cempty = eA(1-f)/d

C(f) = KeAf/d + eA(1-f)/d

my original idea was to write KC_0 + C_0(1-f), would this be the correct way of writing this equation as a function of f?
 
  • #4
I'm not certain what you are looking for here. You are missing an f in your last expression (needs to be

KC_0 f + C_0(1-f)

Did they just want an expression for the total capacitance with f in it? Or something else?

If you still have questions, could you post the entire problem?
 
  • #5
alphysicist said:
I'm not certain what you are looking for here. You are missing an f in your last expression (needs to be

KC_0 f + C_0(1-f)

Did they just want an expression for the total capacitance with f in it? Or something else?

If you still have questions, could you post the entire problem?

they wanted Capacitance as a function of f, here's the question:

Consider a parallel-plate capacitor that is partially filled with a dielectric of dielectric constant K. The dielectric has the same same height as the separation of the plates of the capacitor but fills a fraction f of the area of the capacitor. The capacitance of the capacitor when the dielectric is completely removed is C_0, What is the capacitance C(f) of this capacitor as a function of f?

so as I said it was the Cdielectric + Cempty, so I think what you put will be correct, I'll check it.
 
  • #6
it was correct, thanks alphysicist!
 
  • #7


alphysicist said:
Hi clope023,

They want Ch=Cv, so you can set the right hand sides equal to each other:

2KAe/d(K+1) = Ae(1-f+Kf)/d

and solve for f. When you solve for f, all of the other unknowns will cancel except for k.

...I need help...
My algebra seems to have left me; I can't get the right answer whilst trying to salve for f...
 
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FAQ: Capacitors with partial dielectrics

What are partial dielectrics in capacitors?

Partial dielectrics in capacitors refer to a situation where only a portion of the space between the capacitor's plates is filled with a dielectric material. This results in a variable capacitance depending on the area of the plates that are covered by the dielectric.

How do partial dielectrics affect the capacitance of a capacitor?

Partial dielectrics can increase or decrease the capacitance of a capacitor depending on the dielectric constant of the material used and the amount of area covered by the dielectric. In general, a higher dielectric constant and larger area of coverage will result in a higher capacitance.

What are some common applications of capacitors with partial dielectrics?

Capacitors with partial dielectrics are commonly used in tuning circuits, where the capacitance can be adjusted by changing the area of the plates that are covered by the dielectric. They are also used in variable capacitors for radio frequency applications.

What types of dielectric materials are commonly used in capacitors with partial dielectrics?

Some common dielectric materials used in capacitors with partial dielectrics include air, mica, ceramic, and plastic films such as polyester and polypropylene. The choice of dielectric material depends on the specific application and desired capacitance range.

How do partial dielectrics affect the stability of a capacitor?

Partial dielectrics can decrease the stability of a capacitor due to the variable capacitance resulting from changes in the area of coverage by the dielectric. This can cause fluctuations in the capacitance and impact the performance of the circuit. Special care must be taken in the design and selection of materials to ensure stability in these types of capacitors.

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