Two red blood cells each have a mass of 9.05 x 10 -14 kg and carry a negative ch

Question

Two red blood cells each have a mass of 9.05 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. One cell carries -2.50 pC and the other -3.50 pC, and each cell can be modeled as a sphere 3.75 x 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed, what initial speed would each need so that they get close enough to just barely touch? Assume that there is no viscous drag from any of the surrouding liquid. Also, what is the maxiumum acceleration of the cells as they move toward each other and just barely touch?

Explanation / Answer

# Given

Mass of blood cells m = 9.05 x 10-34 kg

Charge of first blood cell q1 = -2.50 pC = -2.50 x 10-12 C

Charge of second blood cell q2 = -3.50 pC = -3.50 x 10-12 C

Radius of the of the sphere they can be molded into r = 3.75 x 10-6 m

# Solution

1)

The minimum distance between the charges can be the diameter of the sphere they can be molded into

So D = 2r = 2 x 3.75 x 10-6

D = 7.5 x 10-6 m

The potential energy of the blood cells when the distance between them is minimum

E = kq1q2/D

E = 9 x 109 x -2.50 x 10-12 x -3.50 x 10-12 / 7.5 x 10-6

E = 10.5 x 10-9 J

Since the blood cells have same mass this energy is shared between them equally

So the potential energy of one cell is E1 = E2 = E/2 = 5.25 x 10-9 J

This much be the kinetic energy of the blood cells when they are far apart

So E1 = ½ mv2

5.25 x 10-9 = ½ x 9.05 x 10-14 x v2

v2 = 116022.09945

v = 340.62 m/s

2)

The maximum force of repulsion will be experienced by the charges when the distance between them is minimum

F = kq1q2/D2

F = 9 x 109 x -2.50 x 10-12 x -3.50 x 10-12 / (7.5 x 10-6)2

F = 78.75 x 10-15 / 56.25 x 10-12

F = 1.4 x 10-4 N

Maximim acceleration a = maximum force F/ mass m

a = 1.4 x 10-3 / 9.05 x 10-14

a = 1.55 x 1010 m/s2

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