Two red blood cells each have a mass of 4.60×10−14 kg4.60×10−14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. Once cell carries −2.00 pC−2.00 pC of charge and the other −2.90 pC−2.90 pC , and each cell can be modeled as a sphere 8.20 μm8.20 μm in diameter. What minimum relative speed vv would the red blood cells need when very far away from each other to get close enough to just touch? Ignore viscous drag from the surrounding liquid.

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moya1316 moya1316 · Answer 1 · 2020-09-24T05:13:55+02:00

Answer:

v = 5.26 10² m / s

Explanation:

We can solve this exercise using the concepts of conservation of mechanical energy, because there is no friction

starting point. Red blood cells too far away

Em₀ = K = ½ m v²

final point. Red blood cells touching r = 8.20 10⁻⁶ m

Em_f = U = k q₁ q₂ / r₁₂

Em₀ = Em_f

½ m v² = k q₁ q₂ / r₁₂

v = √ (2 k q₁ q₂ / m r₁₂)

we calculate

v =√ (2 9 10⁹ 2 10⁻¹² 2.9 10⁻¹² / (4.60 10⁻¹⁴ 8.20 10⁻⁶))

v = √ (0.276775 10⁶)

v = 0.526 10³ m / s

v = 5.26 10² m / s