Calculation involving Parallel Plate Capacitor


by popsune
Tags: calculation, capacitor, involving, parallel, plate
popsune
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#1
Jun2-06, 01:12 PM
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Hi, I've a problem with this question and I'm not sure if I've worked it out correctly.

This is the question:

The membrane that surrounds a certain type of living cell has a surface area of 5.0 mm2 and a thickness of 10 nm. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.0. If the potential inside the membrane is -60.0 mV with respect to the outside how much charge resides on the outer surface? If the charge is due to monovalent ions, how many such ions are present on the outer surface?
Epsilon0 = 8.85 X 10-12 C2/(N.m2)
e = 1.6 X 10-19 C

This is my working:

V=Q/Ak(Epsilon0)d
-60X10^-3=Q/(5.0X10^-6)X5.0X(8.85X10^-12)X(10X10^-9)
Q=-1.3275X10^-25C

Charge of 1 electron=1.6X10^-19C
No of ions = 1.3275X10-25/1.6X10^-19=8.296X10^-7
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berkeman
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#2
Jun3-06, 01:24 AM
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Welcome to PF, popsune. Keep in mind that C = epslon * A / d as an approximation that matches your homework problem, and that epsilon = epsilon0 * epsilonR
popsune
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#3
Jun3-06, 03:05 AM
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Quote Quote by berkeman
Welcome to PF, popsune. Keep in mind that C = epslon * A / d as an approximation that matches your homework problem, and that epsilon = epsilon0 * epsilonR
Hi thanks for the reply. But can you elaborate on your explanation, cos I don't really get it.

arunbg
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#4
Jun3-06, 07:06 AM
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Calculation involving Parallel Plate Capacitor


What Berkeman is trying to say is that your expression for V is wrong .
V = Q/C right ?

Now for a parallel plate capacitor ,
[tex]C = \frac{\epsilon_0\epsilon_rA}{d}[/tex]

So Q/C becomes ....
popsune
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#5
Jun3-06, 07:21 AM
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Quote Quote by arunbg
What Berkeman is trying to say is that your expression for V is wrong .
V = Q/C right ?

Now for a parallel plate capacitor ,
[tex]C = \frac{\epsilon_0\epsilon_rA}{d}[/tex]

So Q/C becomes ....
Thanks, but what is Epsilon R?
Hootenanny
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#6
Jun3-06, 12:39 PM
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[itex]\epsilon_{r}[/itex] is the relative permitivity of the dielectric medium between the plates. A dielectric material is one which contains polar molecules. The polarity of the molecules will be randomly orientate when no electric field is applied. However, when an electric field is applies the material becomes polarised (each polar molecule is arranged the same), this effect decreases the effective electric field. However, as the electric field strength in inversly proportional to the capacitance, the capacitance increases due to these dipole moments (magnitude of charge multiplied by the distance between them). This is only a brief outline but I'm sure google could reveal more.

~H


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