Question

Two red blood cells each have a mass of 9.05×10−14 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 −3.30 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.

Answer 1

Answer:

v = 330.7 m/s

Explanation:

Let say initially the speed of two blood cells is v and they are far apart from each other

when they come to the closest distance to each other then in that case the initial total kinetic energy of two blood cells will convert into electrostatic potential energy of the cells

So we can say

$\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = \frac{kq_1q_2}{2r}$

here we know that

$m = 9.05 \times 10^{-14} kg$

$q_1 = -2.50 pC$

$q_2 = -3.30 pC$

$r = 3.75 \times 10^{-6} m$

now we will have

$(9.05 \times 10^{-14})v^2 = \frac{(9 \times 10^9)(2.50 \times 10^{-12})(3.30 \times 10^{-12})}{2(3.75 \times 10^{-6})}$

$v^2 = 1.09 \times 10^5$

$v = 330.7 m/s$