Question

A red blood cell may carry an excess charge of about -2.5××10−12−12 C distributed uniformly over its surface. The cells, modeled as spheres, are approximately 6.6 μμm in diameter and have a mass of 9.0××10−14−14 kg.

What is the surface charge density σσ on the red blood cell? Express your answer in C/m22. (Express your answer to two significant figures.)

Cumulative Problem 4 A red blood cell may carry an excess charge of about-2.5×10-12 C distributed uniformly over its surface. The cells modeled as spheres, are approximately 6.6 μ m in diameter and have a mass of 9.0×10-14 kg 1) How many excess electrons does a typical red blood cell carry? (Express your answer to two significant figures.) 1.56 x10" electrons Submit You currently have 2 submissions for this question. Only 10 submission are allowed You can make 8 more submissions for this question 2) 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. (Express your answer to two significant figures.) 1.58 x10-10 Submit You currently have 1 submissions for this question. Only 10 submission are allowed You can make 9 more submissions for this question 3) What is the surface charge density ơ on the red blood cell? Express your answer in C/m2 . (Express your answer to two significant figures.) 1.24 x10-2 C You currently have 2 submissions for this question. Only 10 submission are allowed You can make 8 more submissions for this question Submit m2

Answer 1

1.

Relation between charge and number of electrons is given by:

Q = ne

n = Q/e

e = charge on single electron = -1.6*10^-19

Q = -2.5*10^-12 C

So,

n = Q/e

n = (-2.5*10^-12 C)/(-1.6*10^-19 C)

n = 1.6*10^7 electrons  

(You need to enter your final answer in two significant figures, because given values in question have only two significant figure)

2.

mass of single electron, Me = 9.1*10^-31 kg

mass of cells, Mc = 9.0*10^-14 kg

So,

mass of excess electrons = n*Me

Now we need to calculate

n*Me/Mc = 1.6*10^7*9.1*10^-31/(9.0*10^-14)

ratio = 1.6*10^-10

(Again your final answer should be in two significant figures.)

3.

Surface charge density is given by:

$\sigma$ = Q/A

A = surface Area of red cell = 4*pi*r^2 = pi*d^2

where d = diameter = 6.6*10^-6 m

$\sigma$ = (-2.5*10^-12)/(pi*(6.6*10^-6)^2)

$\sigma$ = -0.018

$\sigma$ = -1.8*10^-2 C/m^2

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