Question

Two red blood cells each have a mass of
9.05
×
10
−
14
k
g
and carry a negative charge spread uniformly over their surfaces.The repulsion arising from the excess charge prevents the cells from dumping together.One celll carries -2.1 uC and the other -3.1 uC and each cell can be modeled as a sphere
3.75
×
10
−
4
m
in radius. If the red blood cells store 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?​

Answer 1

Answer: 29.367 $\frac{m}{s}$

Explanation:

Given data:

The mass of each blood cell =  m = 9.05 X $10^{-14}$ kg.

The charge on one cell = $q_{1}$ = -2.1 uC = -2.1 X $10^{-12}$ C.

The charge of second cell = = $q_{2}$ = -3.1 uC = -3.1 X $10^{-12}$ C.

The radius of the sphere = r = 3.75 X $10^{-4}$ m.

Acceleration is given by,

ma = $q_{1}$ $q_{2}$  / $4\pi e_0(2r^{2} )$  ......................... equation 1

Applying law of conservation of energy to the above expression,

Potential energy + kinetic energy = constant

$q_{1}$ $q_{2}$  / $2\pi e_0r$ + $\frac{1mv^2}{2}$ = constant  ................ equation 2

Let the initial distance be ∞

Let initial speed be $v_1$

Final distance =  3.75 X $10^{-4}$ m

Final speed $v_2$ will be 0

Applying the above data in equation 2 , equation 2 can be rewritten as

$q_{1}$ $q_{2}$  / $2\pi e_0$(∞) + 2(  $\frac{1mv^2}{2}$  ) = $q_{1}$ $q_{2}$  / $4\pi e_0(2r )$  +  0

$mv^2}$ = $q_{1}$ $q_{2}$  / $4\pi e_0(2r )$   ....................equation 3

Substituting values of equation 1 & 2 in equation 3,

9.05 X $10^{-14}$  $v^2$ =  (-2.1 X $10^{-12}$ ) ( -3.1 X $10^{-12}$ )  /  4π (8.85 X $10^{-12}$)(( 2 X 3.75 X $10^{-4}$ )

$v^2$ = 6.51 X $10^{-24}$  /   (8.34 X $10^{-14}$) ( 9.05 X $10^{-14}$ )

$v^2$ =  6.51 X $10^{-24}$  / 7.55 X$10^{-27}$

$v^2$  = 862.418

$v$  = 29.367 $\frac{m}{s}$

v = 29.367 $\frac{m}{s}$

So, initial speed would each need so that they get close enough to just barely touch will be 29.367 $\frac{m}{s}$ .

Answer 2

Thus the speed of the red blood cells is v = 9.27 m/s

Explanation:

Total kinetic energy of two blood cells will convert into electrostatic potential energy of the cells

So we can say  that:

1 / 2 mv^2 + 1 / 2 mv^2 = kq1q2 / 2r

We know that

m = 9.05 x 10^-14 kg

q1 = -2.1 μC

q2 = -3.1 μC

r = 3.75 x 10^-4 m

( 9.05 x 10^-14 ) v^2 = (9 x 10^9) (- 2.1 x 10^-6 ) (3.1 x 10^-6) / 2 x 3.75 x 10^-14

v^2 = 58.59 x 10^9-6-6 / 7.5  10^-14 x 9.05 x 10^-14

v^2 = 58.59 x 10^25 / 7.5 x 9.05

v^2 = 58.59 x 10^25 / 67.875

v^2 = 0.86 x 10^25

v = 9.27 m/s

Thus the speed of the red blood cells is v = 9.27 m/s