How Does Inserting Different Dielectric Materials Affect Capacitor Performance?

In summary: So the question is still incomplete.In summary, a parallel plate condenser with a charge of Q has a plate of thickness t1 and dielectric constant k1 placed between its plates. The rest of the space has another plate of thickness t2 and dielectric constant k2. The potential difference across the condenser is equivalent to the series combination of two capacitors with capacitances C1 and C2, where C1 is the capacitance of the first plate with thickness t1 and dielectric constant k1, and C2 is the capacitance of the second plate with thickness t2 and dielectric constant k2. It is not specified if t1 and t2 are equal, so the question is incomplete.
  • #1
Abhimessi10
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0

Homework Statement


Between the plates of parallel plate condenser having charge Q,a plate of thickness t1 and dielectric constant k1 is placed.In the rest of the space,there is another plate of thickness t2 and dielectric constant K2.The potential difference across the condenser will be.

Homework Equations


C=kAe0/(d-t+t/k)
k----->dielectric constant
A--->Area of the plates
e0---->permittivity
d---->distance between the plates
t------>thickness of the plate

V=Qnet/C

The Attempt at a Solution



Combination of capacitors with materials of dielectric constant K1 and K2 is obtained.i am not sure if this parallel or series connection.

My logic says they are in series.

So Cnet=C1C2/C1+C2

But i am not arriving at a final answer

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  • #2
Your logic is correct, they are in series. Also your formula for the equivalent capacitance is correct. So now what are C1 and C2?
 
  • #3
kuruman said:
Your logic is correct, they are in series. Also your formula for the equivalent capacitance is correct. So now what are C1 and C2?

Yeah but isn't the distance between the plates of C1 d/2 ?
 
  • #4
Yeah but isn't the distance between the plates of C1 d/2 ?
Not on my reading of the Q.
Abhimessi10 said:
... ,a plate of thickness t1 and dielectric constant k1 is placed. In the rest of the space,there is another plate of thickness t2 and ...
d---->distance between the plates
t------>thickness of the plate
the dielectric plates are thickness t1 and t2 and they filled the gap d between the plates, so that t1 + t2 = d
It is not stated that t1 = t2
 
  • #5
IMHO, the question is both incomplete and ambiguous but, given the information provided, can only be solved as 'C in Series'.
 
  • #6
Nik_2213 said:
IMHO, the question is both incomplete and ambiguous but, given the information provided, can only be solved as 'C in Series'.

Is there any combination in which the area of both plates become A/2(not related to this question)
 
  • #7
You can have the gap in the bottom half of the capacitor be completely filled with one kind of dielectric and the top half filled with a different kind. Is that what you mean? (Not the case here).
 
  • #8
kuruman said:
You can have the gap in the bottom half of the capacitor be completely filled with one kind of dielectric and the top half filled with a different kind. Is that what you mean? (Not the case here).
I thought that's exactly what we have here? EDIT: Except that it's not halves
capacitor_double.png
 

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  • #9
Merlin3189 said:
I thought that's exactly what we have here? EDIT: Except that it's not halves
View attachment 230773
That's exactly my interpretation.
 
  • #10
Abhimessi10 said:
Is there any combination in which the area of both plates become A/2(not related to this question)
Only if the dielectric slabs are arranged side by side and so divide the plate area between them. The width of the dielectrics would have to be identical in order to divide the area of the plates equally.

upload_2018-9-16_19-41-48.png


In this case you could consider the resulting device to be two capacitors with the same plate area (A/2) but different dielectrics and connected in parallel.
 

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Related to How Does Inserting Different Dielectric Materials Affect Capacitor Performance?

1. How do capacitors combine in series?

In series, capacitors combine by adding their capacitances together. This means that the total capacitance of the combination is equal to the sum of the individual capacitances. The total voltage across the combination is equal to the sum of the individual voltages.

2. What is the formula for calculating the total capacitance of capacitors in parallel?

The formula for calculating the total capacitance of capacitors in parallel is Ctotal = C1 + C2 + C3 + ..., where C1, C2, C3, etc. are the individual capacitances.

3. How do capacitors combine in parallel?

In parallel, capacitors combine by adding their inverse capacitances together and then taking the inverse of the sum. This means that the total capacitance of the combination is equal to the reciprocal of the sum of the individual inverse capacitances. The total voltage across the combination is equal to the voltage of any individual capacitor.

4. What is the equivalent capacitance of two capacitors with equal capacitances in parallel?

The equivalent capacitance of two capacitors with equal capacitances in parallel is equal to the capacitance of one of the capacitors. This is because in parallel, the voltage across each capacitor is the same, so the total voltage across the combination is equal to the voltage of one of the capacitors. Therefore, the equivalent capacitance is equal to the capacitance of one of the capacitors.

5. Can capacitors be combined in both series and parallel in the same circuit?

Yes, capacitors can be combined in both series and parallel in the same circuit. This can be seen in more complex circuits where there are both series and parallel combinations of capacitors. In these cases, the equivalent capacitance of the entire combination can be calculated by finding the equivalent capacitance of each individual combination and then combining those values using the appropriate formula.

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