Calculation involving Parallel Plate Capacitor

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Homework Help Overview

The discussion revolves around a problem involving a parallel plate capacitor modeled by a biological cell membrane. The original poster presents a calculation to determine the charge on the outer surface of the membrane based on given parameters, including surface area, thickness, dielectric constant, and potential difference.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the formula for voltage in a capacitor to find the charge, but some participants question the validity of their expression for voltage. Others suggest clarifying the relationship between capacitance and charge.

Discussion Status

Participants are actively engaging with the original poster's calculations, offering clarifications on the formulas used and discussing the concept of relative permittivity. There is an ongoing exploration of the underlying principles without a clear consensus on the correctness of the initial approach.

Contextual Notes

Some participants note the need for further elaboration on specific terms, such as the relative permittivity, and the original poster expresses uncertainty about the explanations provided.

popsune
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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
 
Last edited:
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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
 
berkeman said:
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.
 
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 ...
 
arunbg said:
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?
 
[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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