Parallel-plate capacitor with 2 dielectrics?

[Destroxia]

Homework Statement

[/B]
The figure shows a parallel-plate capacitor with a plate area A = 5.29 cm^2 and plate separation d = 8.10 mm. The left half of the gap is filled with material of dielectric constant κ1 = 6.50; the right half is filled with material of dielectric constant κ2 = 11.4. What is the capacitance?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c25/fig25_48_wiley.gif

(Given)

A = 5.29 cm^2
d = 8.10 mm
k1 = 6.50
k2 = 11.4
e₀ = 8.85E-12

(unknowns)

C_{NoDielectrics} = ?
C_Dielectrics = ?

Homework Equations

C[/B]_{Parallel-Plate} = (e₀*A)/d

C_{AddedDielectric} = C_{Parallel-Plate}*K_{dielectricconstant}

The Attempt at a Solution

C_{No Dielectric}/2 = (e₀*((5.29E-2)/2))/(8.10E-3) = 2.89E-11

Since the separated capacitor is in series (I think), you can use the law of inverse capcitance sums, after multiplying the C_{No Dielectric}/2 once by k1, and another time by k2, and add them, then inverse them, you get

1/((C_{No Dielectric}/2)*k1) + 1/((C_{No Dielectric}/2)*k2) =

= 3.04E-11 + 5.32E-11 = 8.36E-11

1/8.36E-11 = 1.20E-12 = C_Dielectrics

 

[Destroxia]

Apparently, my image couldn't be posted as it was a secure URL. I will provide a description. It is 1 parallel-plate capacitor split in half VERTICALLY, each half has a different dielectric. The one on the left is K1, the one of the right is K2. The Area value is for the whole parallel plate capacitor, and each half is A/2. d is the distance between plates.

 

[gneill]

Apparently, my image couldn't be posted as it was a secure URL. I will provide a description. It is 1 parallel-plate capacitor split in half VERTICALLY, each half has a different dielectric. The one on the left is K1, the one of the right is K2. The Area value is for the whole parallel plate capacitor, and each half is A/2. d is the distance between plates.

What is your logic for considering the two capacitances to be in series?

 

[Destroxia]

What is your logic for considering the two capacitances to be in series?

I figured that it would just be because 1 came after the other, but I thought about it for awhile and came to the conclusion that isn't it just the voltage remaining constant throughout, or the charge remaining constant which determines they are in series, or parallel? So, would these actually be in parallel, since it is "Technically" the same capacitor, just with different dielectrics, then the voltage must be constant? Implying it would be parallel... ?

 

[gneill]

They are parallel because they share the same two nodes at their connection points to any circuit: they share the same wire leads. That also means of course that they share the same potential difference.

So re-work your calculation with the assumption that the two sections are in parallel.

 

[Destroxia]

They are parallel because they share the same two nodes at their connection points to any circuit: they share the same wire leads. That also means of course that they share the same potential difference.

So re-work your calculation with the assumption that the two sections are in parallel.

I reworked the calculations for the capacitance, the program is giving me the wrong answer, even though I calculated them in parallel.

And yes, it wants the answer in just Farads.

 

[gneill]

Check your area value. The given units is square centimeters. What's that in square meters?

Otherwise, your method looks fine.

 

[SammyS]

Seems like the image finally showed up for me.

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c25/fig25_48_wiley.gif

 

[Destroxia]

Check your area value. The given units is square centimeters. What's that in square meters?

Otherwise, your method looks fine.

OH GOODNESS! I can't believe I overlooked that, It's always the units that get me. Thank you so much!