Question

Two red blood cells each have a mass of 9.0××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 -2.20 pC of charge and the other -3.30 pC, and each cell can be modeled as a sphere 7.5 μμm in diameter.

1) What speed would they need when very far away from each other to get close enough to just touch? Assume that there is no viscous drag from any of the surrounding liquid.

2) What is the maximum acceleration of the cells in the previous part?

Answer 1

1) Applying energy conservation,

PEi + KEi = PEf+ KEf

{ electric PE = k q1 q2 / r when they are very far apart then PE = 0 }

0 + 2 ( m v^2 / 2) = [ (9 x 10^9 x - 2.20 x 10^-12 x - 3.30 x 10^-12) / (7.5 x 10^-6)] + 0

(9 x 10^-14) v^2 = 8.712 x 10^-9

v = 311 m/s ........Ans

2) maximum acc will be when force is maximum that will when they are touching,

Fe = k q1 q2 / (r + r)^2 = k q1 q2 / d^2

a = Fe / m

a = (9 x 10^9 x 2.20 x 10^-12 x 3.30 x 10^-12) / ( (7.5 x 10^-6)^2 (9 x 10^-14))

a = 1.29 x 10^10 m/s^2 .....Ans