Red blood cells often carry an electrical charge. Consider two red blood cells w

Question

Red blood cells often carry an electrical charge. Consider two red blood cells with the following charges: 21.2 pC and +54.4 pC. The red blood cells are 2.94 cm apart. (1 pC = 1 1012 C.)

(a) What is the magnitude of the force on each red blood cell? N Are the red blood cells attracted or repulsed by each other? attracted repulsed

(b) The red blood cells come into contact with each other and then are separated by 2.94 cm. What magnitude of force does each of the red blood cells now experience? N Are the red blood cells attracted or repulsed by each other?d

Explanation / Answer

a) The forCe between charges is given by Coulomb's law as KqQ/r^2

F = 9× 10^9 x 21.2 x 10^-12 x 54.4 x 10^-12 / (0.0294)^2

Or, F= 1.2008 x 10^-8 N attractive force as unlike charges attract each other

b) Once the cells are brought in contact, their charge gets uniformly redistributed. Thus, each cell now contains (54.4-21.2)/2 = + 16.6 pC.

Now since each cell has a positive charge, they repel each other. F= 9×10^9 x 16.6 x 10^-12 x 16.6 x 10^-12/ (.0294)^2

F= 2.869 x 10^-9 N repulsive force.

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