Answer:
Explanation:
Given that :
Mass of the red blood cell m= [tex]1.00*10^{-12} \ kg\\[/tex]
Density of the blood P = 1100 kg/m³
Dielectric constant K = 5.00
Potential Difference [tex]\Delta V = 100 mV[/tex] = 100×10⁻³ V
Thickness d = 95 nm = 95 × 10⁻⁹ m
Permittivity of free space [tex]\epsilon_o[/tex] = [tex]8.85 *10^{-12} C^2/N.m^2[/tex]
a)
the volume of the cell and thus find its surface area.
The volume of the cell is being calculated as follows:
[tex]V = \frac{m}{\rho} \\ \\ = \frac{1.00*10^{-12}}{1100 } \\ \\ = 9.09*10^{-16} m^3[/tex]
Here the model of the red blood cell depicts that the cell is a spherical capacitor. Thus , the volume of the sphere is:
[tex]V = \frac{4}{3} \pi r^3 \\ \\ r^3 = \frac{3V}{4 \pi } \\ \\ r = (\frac{3V}{4 \pi})^{1/3}[/tex]
The radius of the sphere is = [tex][\frac{3(9.09*10^{-14} m^3)}{4 \pi} ]^{(1/3)[/tex]
= 6.01654×10⁻⁴ m
The surface area of the cell is A = 4 πr²
= 4 π ( 6.01654×10⁻⁴)²
= 4.548×10⁻¹⁸ m²
b) Estimate the capacitance of the cell.______ F
the capacitance of the cell ;
C = [tex]\frac{ \epsilon_o *A}{d}[/tex]
C = [tex]\frac{(5.00)(8.85*10^{-12}*4.548*10^{-10}}{95*10^{-9}}[/tex]
C = 2.11 × 10⁻¹³ F
c) Calculate the charge on the surface of the membrane._________ C
How many electronic (elementary) charges does the surface charge represent? _____
The charge on the surface of the membrane is given by:
[tex]Q = C( \Delta V) \\ \\ Q =2.11*10^{-13} *100*10^{-3} \\ \\ Q = 2.11*10^{-14} C[/tex]
the numbers of electrons in the surface of the charge is :
[tex]n = \frac{Q}{e} \\ \\ where\ \ e = .602*`10^{-19} \ C \\ \\ n = \frac{2.11*10^{-14}C}{1.602*10^{-19}}[/tex]
[tex]n = 1.32*10^5 \ \ \ electrons[/tex]
