(a) If an average red blood cell has a mass of 1.10 10 12 kg, estimate the volum

Question

(a) If an average red blood cell has a mass of 1.10 1012 kg, estimate the volume of the cell and thus find its surface area. The density of blood is 1,100 kg/m3. (Assume the volume of blood due to components other than red blood cells is negligible.)

(b) Estimate the capacitance of the cell by assuming the membrane surfaces act as parallel plates.
F

(c) Calculate the charge on the surface of the membrane.
C

How many electronic charges does the surface charge represent?

volume     m3 surface area     m2 5 points SCrCF D 16 P.053. Oi4 Submissioes Used My Notes Ask Your model of red blood cel portrays the cell ss sphCrical capacitor, positivcly ch roca liquid :pncre of surfacc area scparated rom the surrounding nco tivc y charged fluid by a mcmbranc cf thickncss t TIny coct odes i troduced into tho terior of the cell show potential difference of 100 mv cross the membrane. The membranc's thickness is estimated to bc 103 nm and has dielectric constant of 5.00 a) an averaga rad hicod call haa a mass at 1.10x 10-12 kg, astimata tha volume at tha call and thus tind its surtace area. Tha dansity at tload is 1,100 kgim3. (Assuma tha voluma at hlood dua to components ather than rad hlcod calls ia nagligihbia surface aree m2 b) Est mate tha capactance of the call by assuming the membrane surfaces act as parallal plates. c) Calculate the charge an the surtaa ot tha mamhrana. Need Help? Sutbmit Anewer Sawe Progress

Explanation / Answer

Part A)

Volume is found via the density formula = m/V

S0 V = m/

V = (1.1X 10-12)/1100

V = 10-15 m3

Using that volume, we can find the radius of the blood cell sphere

The formula for the Volume of a sphere is V = 4/3(r3)

Solve for r

r = (3V/4)1/3

r = 6.205 X 10--6 m

Surface area of a sphere is found by the formula A = 4r2

So A = 4()(6.205 X 10-6)2

A = 4.835 X 10-10 m2

Part B)

For capacitance apply the formula C = kA/d

k = 5 and = 8.85 X 10-12 C2/Nm2

d = 103 nm

C = (5)(8.85 X 10-12)(4.835X 10-10)/(103 X 10-9)

C = 2.08 X 10-13 F

Part C)

To find charge, we use Q = CV

Q = (2.08X 10-13)(100 X 10-3)

Q = 2.08X 10-14 C

part d)

To find the number of charges that involves, we know that one elememtal charge has a value of

1.6 X 10-19 C

Therefore the number of charges will be total charge divided by elemental charge

2.08 X 10-14/1.6 X 10-19

Total number of charges is 1.298 X 105

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