Register to reply 
Calculation involving Parallel Plate Capacitor 
Share this thread: 
#1
Jun206, 01:12 PM

P: 3

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 1012 C2/(N.m2) e = 1.6 X 1019 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 = 1.3275X1025/1.6X10^19=8.296X10^7 


#2
Jun306, 01:24 AM

Mentor
P: 41,109

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
Jun306, 03:05 AM

P: 3




#4
Jun306, 07:06 AM

P: 600

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 .... 


#5
Jun306, 07:21 AM

P: 3




#6
Jun306, 12:39 PM

Emeritus
Sci Advisor
PF Gold
P: 9,772

[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 


Register to reply 
Related Discussions  
Nonparallel Plate Capacitor  Introductory Physics Homework  2  
A proton is released from rest at the positive plate of a parallelplate capacitor.  Introductory Physics Homework  16  
Parallel Plate Capacitor  Introductory Physics Homework  1  
ParallelPlate Capacitor  Introductory Physics Homework  7  
Parallel plate capacitor  Introductory Physics Homework  3 