Calculation involving Parallel Plate Capacitor

In summary, the conversation is about a problem involving a living cell membrane acting as a parallel plate capacitor with a dielectric constant of 5.0. The potential inside the membrane is -60.0 mV with respect to the outside, and the question asks for the amount of charge on the outer surface and the number of monovalent ions present. The working provided calculates the charge to be -1.3275X10^-25C and the number of ions to be 8.296X10^-7. However, the expression for V is incorrect and should be V = Q/C, where C = \frac{\epsilon_0\epsilon_rA}{d}, and \epsilon_r is the relative permitivity of
  • #1
popsune
3
0
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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  • #2
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
 
  • #3
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.
 
  • #4
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 ...
 
  • #5
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?
 
  • #6
[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
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy by creating a potential difference between two conductive plates separated by a dielectric material.

2. How do you calculate the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

3. What is the relationship between capacitance and plate area in a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the area of the plates. This means that as the plate area increases, the capacitance also increases.

4. How does the distance between plates affect the capacitance of a parallel plate capacitor?

The distance between plates is inversely proportional to the capacitance of a parallel plate capacitor. This means that as the distance between plates increases, the capacitance decreases.

5. How can the capacitance of a parallel plate capacitor be increased?

The capacitance of a parallel plate capacitor can be increased by increasing the area of the plates, decreasing the distance between plates, or using a dielectric material with a higher permittivity. Additionally, connecting multiple parallel plate capacitors in series or in parallel can also increase the overall capacitance.

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