A model of a red blood cell portrays the cell as a spherical capacitor, a positi

Question

A model of a red blood cell portrays the cell as a spherical capacitor, a positively charged liquid sphere of surface area A separated from the surrounding negatively charged fluid by a membrane of thickness t. Tiny electrodes introduced into the interior of the cell show a potential difference of 100 mV across the membrane. The membrane's thickness is estimated to be 98 nm and has a dielectric constant of 5.00.

(a) If an average red blood cell has a mass of 1.2e-12 kg, estimate the volume of the cell and thus find its surface area. The density of blood is 1100 kg/m3.

volume = m3 ?

surface area = m2?

(b) Estimate the capacitance of the cell by assuming the membrane surfaces act as parallel plates. F ?

(c) Calculate the charge on the surface of the membrane. C ?

How many electronic (elementary) charges does the surface charge represent?

Explanation / Answer

here,

potential difference, V = 100 mV = 0.1 V
thickness of membrane, t = 98 nm = 98*10^-9 m
dielectric constant, k = 5

Part A:
mass , m = 1.2*10^-12 kg
density, p = 1100 kg/m^2

Volume = mass/density = 1.2*10^-12/ 1100 = 1.091*10^-15 m^3

Since, V = 4/3*pi*r^3
4/3*pi*r^3 = 1.091*10^-15
r = 6.386*10^-6 m ( Radius of cell)

Therefore Surface Area, A = 4/3*pi*r^2
A = 4/3*pi*(6.386*10^-6)^2
A = 1.708*10^-10 m^2

Part B:
Capacitance, C = k*eo*A/t
C = 5 * 8.85*10^-12*(1.708*10^-10)/(98*10^-9)
C = 7.712*10^-14 F

Part C:
SInce, Charge,
Q = C*V
Q = 7.712*10^-14 * 0.1
Q = 7.712*10^-15 C

Therefore elementary charge, Sigma = Charge/A = 7.712*10^-15 / (1.708*10^-10)

Sigma = 4.515*10^-5 C/m^2

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