Question

Two red blood cells each have a mass of 9.0 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion from the excess charge prevents the cells from clumping together. One cell carries -2.5pC 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.
1. What initial speed would each need so that they get close enough to just barely touch?
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?

Answer 1

Answer:

Explanation:

Given that:

The mass of the cell is 9.0 x 10^-14 kg

The charges of the cell is -2.5pC and the other -3.30 pC

$q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C$

Radius is  3.75 × 10-6 m

The final distance is twice the radius

i.e $2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m$

The formula for the velocity of the cell is

$mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r}$

$v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }$

$=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s$

The maximum acceleration of the cells as they move toward each other and just barely touch is

$ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}$

$=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}$

$=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2$

Answer 2

The answers obtained are;

1. The initial speed of each of the red blood cells is $v= 331.66\,m/s$.

2. The maximum acceleration of the cells is $a=1.47\times 10^{10}\,m/s^2$.

The answer is explained as shown below.

We have, the mass of the red blood cell;

• $m=9\times 10^{-14}\,kg$

Also, the charges of the cells are;

• $q_1=-2.5\times 10^{-12}\,C$ and

• $q_2=-3.30\times 10^{-12}\,C$

The distance between the charges when they barely touch will be two times the radius of each charge.

• $r=2\times r\,'=2\times3.75\times10^{-6}\,m=7.5\times10^{-6}\,m$

Kinetic Energy of moving charges

1. As both the cells are negatively charged they will repel each other.

• So, for the cells to come nearly close, their kinetic energies must be equal to the electric potential between them.

• $\frac{1}{2}mv^2+ \frac{1}{2}mv^2=k\frac{q_1 q_2}{r^2}$

• Where, $k=9\times10^9\,Nm^2/C^2$ is the Coulomb's constant.

Now, substituting all the known values in the equation, we get;

• $(9\times 10^{-14}\,kg)\times v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m}$

$v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2$

• $\implies v=\sqrt{110000\,m^2/s^2}=331.66\,m/s$

Electrostatic force between two charges

2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.

• $ma=k\frac{q_1 q_2}{r^2}$

Substituting the known values, we get;

$(9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}$

$\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }$

• $a=1.47\times 10^{10}\,m/s^2$

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