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 -2.60 pC of charge and the other -3.80 pC, and each cell can be modeled as a sphere 7.5 um 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. (Express your answer to two significant figures.) m/s Submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. 2) What is the maximum acceleration of the cells in the previous part? (Express your answer to two significant figures.) x1010 m/s? Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Answer 1

Their potential energy is U = kq₁q₂/d.

At a large distance (r→∞) U = 0 and their initial kinetic energy is K = 2 x ½mv² = mv² (where m is the mass of a single cell).

As they get closer, the slow down due to repulsion: kinetic energy is turned to potential kinetic energy. At the point they just touch, they have momentarily stopped and their potential energy equals whatever their initial kinetic energy was.

mv² = kq₁q₂/d

v = √[kq₁q₂/(md)]

'd' is the distance between their centers when they just touch, this is 2 radii.

I’ll use exponential notation, e.g. 3x10⁸ is written as 3e8.

v = √[kq₁q₂/(md)]

. .= √[8.99e9 x (-2.60e-12) x (-3.80e-12) x/(9.0e-14 x 7.5e-6)]

. .= 363 m/s

__________________________

The max. acceleration of a cell occurs when the force is a maximum. F = kq₁q₂/d². This is when they are closest, just touching (d=2r, as above).

Since F = ma, a = F/m

a_max = (kq₁q₂/(2r)²)/m

. . . . . .= kq₁q₂/(4r²m)

. . . . . .= 8.99e9 x (-2.60e-12) x (-3.80e-12) x/(9.0e-14 x 4 x (7.5e-6)²)

. . . . . .= 4.39e9 m/s²

= 0.439e10 m/s²