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 -2.10 pC and the other 2.60 pc, and each cell can be modeled as a sphere 3.75 × 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 surrounding liquid initial speed: 269.01 m/s What is the maximum acceleration of the cells as they move toward each other and just barely touch? maximum acceleration m/s

Answer 1

From given data finally when touch to each other kinetic energy charges in to potential energy. 1/2 mv^2 + 1/2 mv^2 = kov_1 ov_2/d mv^2 = kov_1ov_2/d v = squareroot kov_1 ov_2/md = squareroot 9 times 10^9 times 2 times 1 times 10^-12 times 2.6 times 10^-12/9.05 times 10^-14 times (2 times 3.75 times 10^-6) v = 269.07 m/sec Since F = ma: a = F/m a_max = kov_1 ov_2/d^2/m = (9 times 10^9 times 2.1 times 10^-12 times 2.6 times 10^-12)/(2 times 3.75 times 10^-6)/9.05 times 10^-14 a_max = 9.65 times 10^9 m/s^2) 2) given that work done w = 1.6 times 10^-20 s we have the relation w = ov increment v, ov = 1.6 times 10^-19 c increment v = w/a = 1.6 times 10^-20 J/1.6 times 10^-19 C increment v = 0.1 v (or) 100 times 10^-3 v increment v = 100 mv