What is the capacity of the capacitor?

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SUMMARY

The capacity of a parallel plate capacitor filled with two dielectric layers can be calculated using the formula C = εA/(d1/ε1 + d2/ε2), where ε1 and ε2 are the relative permittivities of the dielectric materials, and d1 and d2 are their respective thicknesses. The effective thicknesses of the dielectrics are represented as d1/ε1 and d2/ε2. This approach treats the dielectric layers as capacitors in series, requiring careful consideration of the order of operations in calculations. Proper use of parentheses in expressions is essential for clarity and accuracy.

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tsgkl
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Homework Statement


The space between the plates of a parallel plate capacitor is filled consecutively with two dielectric layers of thickness d1 and d2 respectively.If A is the area of each plate,what is the capacity of the capacitor?


Homework Equations


C=\frac{εA}{d}
(for parallel plate capacitor with air as dielectric)

The Attempt at a Solution


i honestly don't know how to attempt this question...can anyone give me some clues...
 
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tsgkl said:

Homework Statement


The space between the plates of a parallel plate capacitor is filled consecutively with two dielectric layers of thickness d1 and d2 respectively.If A is the area of each plate,what is the capacity of the capacitor?


Homework Equations


C=\frac{εA}{d}
(for parallel plate capacitor with air as dielectric)

The Attempt at a Solution


i honestly don't know how to attempt this question...can anyone give me some clues...
Were you given the relative permittivity for each dielectric material?

Hint: Dielectric layers which parallel the plates of the capacitor will effectively behave like capacitors in their own right, carving the original space into a set of capacitors in series. The surface boundaries between materials are the locations of the "plates" of these capacitors. Dielectric surfaces in contact with the conductive plates of the capacitor are considered to be a single plate of one such capacitor (consider the plate to be the electrical contact of that sub-capacitor).
 
nope, relative permittivity for each material ain't given...
 
tsgkl said:
nope, relative permittivity for each material ain't given...

Then you'll have to create your own variable names for them. No big deal since numerical values for the thicknesses and area weren't given either, so a symbolic result is expected.
 
just tell me one thing is effective thickness of dielectric 1/ε the actual thickness?
 
tsgkl said:
just tell me one thing is effective thickness of dielectric 1/ε the actual thickness?

d1 and d2 are the actual thicknesses. I suppose one could consider the effective thicknesses to be ##\frac{1}{ε_1}d_1## and ##\frac{1}{ε_2}d_3##, where ε1 and ε2 are the relative permittivities. Although it is more usual to see the effective permittivity specified as εεo and the distances left alone.
 
as you said the effective thickness to be d11 and d22 so the answer should be εA/d11+d22
 
tsgkl said:
as you said the effective thickness to be d11 and d22 so the answer should be εA/d11+d22

You should use parentheses in your expressions to disambiguate the order of operations. Even so, your expression doesn't have the right form.

How do capacitors in series add?
 

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