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 –2.50 pC of charge and the other –3.10 pC, and each cell can be modeled as a sphere 7.5 μm in diameter. (a) What speed would they need when very far away from each other to get close enough to just touch? Ignore viscous drag from the surrounding liquid. (b) What is the magnitude of the maximum acceleration of each cell in part (a)?

Answer 1

Answer:

321.39 m/s

$1.3772213803 \times 10^{10}\ m/s^2$

Explanation:

m = Mass = $9\times 10^{-14}\ kg$

$q_1$ = Charge = -2.50 pC

$q_2$ = Charge = -3.10 pC

r = Radius = 3.75 μm

$\varepsilon_{0}$ = Vacuum permittivity = $8.85 \times 10^{-12}\ F/m$

We have the relation

$m v^{2} &=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} 2 r}$

$v =\sqrt{\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)(m)}}$

$=\sqrt{\frac{\left(-2.50 \times 10^{-12}\right)\left(-3.1 \times 10^{-12}\right)}{4(3.14)\left(8.85 \times 10^{-12}\right)\left(2 \times 3.75 \times 10^{-6}\right)\left(9 \times 10^{-14}\right)}}$

$=321.39\ m/s$

The Speed they need when very far away from each other to get close enough to just touch is 321.39 m/s

We have the relation

$m a=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)^{2}}$

$a=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)^{2}(m)}$

$=\frac{\left(-2.50 \times 10^{-12}\right)\left(-3.10 \times 10^{-12}\right)}{4(3.14)\left(8.85 \times 10^{-12}\right)\left(2 \times 3.75 \times 10^{-6}\right)^{2}\left(9 \times 10^{-14}\right)}$

$a=1.3772213803 \times 10^{10} \mathrm{m} / \mathrm{s}^{2}$

The acceleration is $1.3772213803 \times 10^{10}\ m/s^2$