Answer:
Step-by-step explanation:
Given that :
Mass of the red blood cell m=

Density of the blood P = 1100 kg/m³
Dielectric constant K = 5.00
Potential Difference
= 100×10⁻³ V
Thickness d = 95 nm = 95 × 10⁻⁹ m
Permittivity of free space
=

a)
the volume of the cell and thus find its surface area.
The volume of the cell is being calculated as follows:

Here the model of the red blood cell depicts that the cell is a spherical capacitor. Thus , the volume of the sphere is:

The radius of the sphere is =
![[(3(9.09*10^(-14) m^3))/(4 \pi) ]^{(1/3)](https://img.qammunity.org/2021/formulas/physics/college/l5bbanp1e4wlu2ipyuqv60rs557yvdsdom.png)
= 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 =

C =

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:

the numbers of electrons in the surface of the charge is :

