Capacitance of a parallel plate capacitor?

Discussion in 'Homework Help' started by smarch, Jun 30, 2010.

1. smarch Thread Starter Active Member

Mar 14, 2009
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A parallel plate capacitor has a surface area of 1 cm2 and its plates are
separated by 3 mm. In between the plates are three layers of dielectric,
each 1 mm thick, with relative permittivities of 3, 5 and 11 respectively. What
is the capacitance of the whole capacitor?

I know I use the formula c=eoerA/d.
Do I add the dielectrics together so er=19? and just apply the formula as usual?

2. t_n_k AAC Fanatic!

Mar 6, 2009
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Treat this as three capacitors (C1, C2 & C3) in series, each with the same physical dimensions (1mm plate separation) but with different relative permittivity [er1=3, er2=5, er3=11].

1/Ctot=1/C1+1/C2+1/C3

Last edited: Jun 30, 2010
3. smarch Thread Starter Active Member

Mar 14, 2009
52
0
c = Aeo/(d1/er1 + d2/er2 + d3/er3)

Is that correct?

4. steveb Senior Member

Jul 3, 2008
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469
That looks correct to me from memory (which gets less reliable as I age).

Based on the other problems you are working on, I recommend that you make sure you can derive this formula from Maxwell's equations. TNK's approach is a good one, but the difficulty you are having with the other problems may go away if you work out this simple case from first principles.

5. KL7AJ AAC Fanatic!

Nov 4, 2008
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Does this really work? I probably ran across this ages ago, but I don't think I've even encountered the question in recent history. Very cool!

Eric

6. t_n_k AAC Fanatic!

Mar 6, 2009
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No ... your equation is incorrect

$C_1=\epsilon_0\epsilon_{r1}\frac{A}{d}$

$C_2=\epsilon_0\epsilon_{r2}\frac{A}{d}$

$C_3=\epsilon_0\epsilon_{r3}\frac{A}{d}$

$\frac{1}{C_{tot}}=\frac{1}{C_1}+ \frac{1}{C_2}+ \frac{1}{C_3}$

or

$C_{tot}=\frac{C_1C_2C_3}{C_1C_2+C_2C_3+C_1C_3}$

or

$C_{tot}=\epsilon_0 \frac{A}{d}$\frac{\epsilon_{r1} \epsilon_{r2} \epsilon_{r3}}{\epsilon_{r1} \epsilon_{r2}+\epsilon_{r2} \epsilon_{r3} + \epsilon_{r1} \epsilon_{r3}}$$

Last edited: Jun 30, 2010
7. steveb Senior Member

Jul 3, 2008
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TNK,

His equation appears to me to just be a different form of your equation.

8. t_n_k AAC Fanatic!

Mar 6, 2009
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Your are quite right. My error. Apologies to 'smarch'.

9. smarch Thread Starter Active Member

Mar 14, 2009
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No worries TNK, I appreciate your help!