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 105 nm and has a dielectric constant of 5.00.
(a) If an average red blood cell has a mass of 1.1e-12 kg, estimate the volume of the cell and thus find its surface area. The density of blood is 1100 kg/m3.

volume = 2.39e-10 m3
**Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.**

surface area = 7.162e-11 m2
**Your response differs from the correct answer by more than 10%. Double check your calculations.**

(b) Estimate the capacitance of the cell by assuming the membrane surfaces act as parallel plates.
3.02e-14 F
**Your response differs from the correct answer by more than 10%. Double check your calculations.**

(c) Calculate the charge on the surface of the membrane.
3.02e-15 C
**Your response differs from the correct answer by more than 10%. Double check your calculations.**

(d) How many electronic (elementary) charges does the surface charge represent?
1.89e+4
**Your response differs from the correct answer by more than 10%. Double check your calculations.**


---Not quite sure where I went wrong. Any help and explanation would be greatly appreciated. Thanks so much!!

Explanation / Answer

a) Volume of the cell is V = m/? = (1.1 x 10^-12 kg)/(1100 kg/m3) = 1.0 x 10^-15 m3 But V = 4/3 p r^3 (1.0 x 10^-15 m3) = 4/3 p r^3 The radius of the cell is r = 6.203 x10^-6 m The surface area is S = 4p r^2 = 4p (6.203 x10^-6 m)^2 = 4.835 x10^-10 m2 b) Thecapacitance of the cell by assuming the membrane surfaces act as parallel plates C =k Aeo/d =k (pr^2)eo/d = (5.00)(p(4.835 x10^-10 m2)^2)(8.85 x 10^-12 C2/N.m2)/(98 x 10^-9 m) =----- F c) The charge on the surface of the membrane is q = CV Here V = 100 x 10^-3 V d) The number of charges are N = q/e Here e =1.6 x 10^-19 C Substitute the values we get the answer.

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