Question

A red blood cell typically carries an excess charge of about −2.5×10−12 C distributed uniformly over its surface. The red blood cells can be modeled as spheres approximately 7.6 μm in diameter and with a mass of 9.0×10−14 kg. How many excess electrons does a typical red blood cell have? excess electrons: 1.56 ×10 7 electrons Does the mass of the extra electrons appreciably affect the mass of the cell? To find out, calculate the ratio of the mass of the extra electrons to the mass of the cell without the excess charge. ratio: 1.58 ×10 −10 What is the surface charge density on the red blood cell? Express your answer in units of C/m2 and electrons/m2. surface change density: −3.44 ×10 3 C/m2 surface charge density: 2.15 ×10 22

Answer 1

Answer:

The number of excess electrons on a red blood cell = $\rm 1.56\times 10^7\ electrons.$

The ratio of the mass of the extra electrons to the mass of the cell without the excess charge = $\rm 1.58\times 10^{-10}.$

The surface charge density on the red blood cell = $\rm -1.38\times 10^{-2}\ C/m^2.$ = $\rm 8.68\times 10^{16}\ electrons/m^2.$

Explanation:

Given:

• Excess charge on the red blood cell, $\rm q=-2.5\times 10^{-12}\ C.$

• Diameter of the red blood cell, $\rm D = 7.6\ \mu C = 7.6\times 10^{-6}\ C.$

• Mass of the red blood cell, $\rm m = 9.0\times 10^{-14}\ kg.$

Finding the number of excess electrons on a red blood cell:

Charge on an electron, $\rm e = -1.6\times 10^{-19}\ C.$

If there are n number of excess electrons on the RBC, then,

$\rm q=ne\\\therefore n = \dfrac qe=\dfrac{-2.5\times 10^{-12}}{-1.6\times 10^{-19}}=1.56\times 10^7\ electrons.$

Calculating the ratio of the mass of the extra electrons to the mass of the cell without the excess charge:

Mass of 1 electron, $\rm m_e=9.11\times 10^{-31}\ kg.$

Mass of n electrons, $\rm M= n\times 9.11\times 10^{-31}=1.56\times 10^7\times 9.11\times 10^{-31}=1.42\times 10^{-23}\ kg.$

The ratio of the mass of the extra electrons to the mass of the cell without the excess charge is given as

$\rm Ratio = \dfrac{M}{m}=\dfrac{1.42\times 10^{-23}}{9.0\times 10^{-14}}=1.58\times 10^{-10}.$

Thus, the mass of the extra electrons does not appreciably affect the mass of the cell.

Calculating the surface charge density on the red blood cell:

It is given that the red blood cells can be modeled as spheres, then, the surface area of the RBC is given as

$\rm A = 4\pi (Radius)^2=4\pi \times \left ( \dfrac D2\right ) ^2=4\pi \times \left ( \dfrac {7.6\times 10^{-6}}2\right ) ^2=1.81\times 10^{-10}\ m^2.$

The surface charge density of the RBC is given as:

$\rm \sigma = \dfrac qA=\dfrac{-2.5\times 10^{-12}}{1.81\times 10^{-10}}=-1.38\times 10^{-2}\ C/m^2.$

In terms of $\rm electrons/m^2$,

$\sigma = \rm \dfrac{n}{A}=\dfrac{1.56\times 10^7}{1.81\times 10^{-10}}=8.68\times 10^{16}\ electrons/m^2.$