Question

5 of 39 Questions Assignment Score: Resources Give Up Feedback Try Again 439/3900 Attempt 3 1 Question 2 of mo Attemots 96/100 Question 5 of 39> Corect Two red blood cells each have a mass of 5.70 x 104 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 of charge and the other -3.50 pC, and each cell can be modeled as a sphere 6.40 um in diameter What minimum relative speed would the red blood cells need when very far away from each other to get close enough to just touch? Ignore viscous drag from thc surrounding liquid 2 Question 2 of oo Attempts Comect 3 Question 100% 1 of oo Attemots corect 4 Question 10096 1 of oo Attempts Correct 4.6x10 m/s 5 Question 4B/100 2 of o Attempts In Frogress What is the magnitude of the maximum acceleration amax of each cell? 0/100 amax= 2.69×1010 m/s 7 Question 0/100 0/100 9 Question 0 of Attempts 0/100

Answer 1

Apply the concept of relative motion and energy conservation ,carefully.

Suppose, cell carries - 2.00 pc of charge is at rest. Hence, minimum relative speed v = Speed of cell carries charge of - 3.50 pc. Hence, we can easily apply energy conservation. Now, for just touch situation. Speed = 0 Distance between them = Sum of radii = 6.40 mu m = 6.40 times 10^-6 m So, Loss in kinetic energy = 1/2 mv^2 - 0 Here, m = mass of cell = 5.70 times 10^-14 kg Gain in electrical potential energy = 1/4 pi epsilon_0 (-2.00 times 10^-12) (-3.50 times 10^-12)/6.40 times 10^-6 = 9 times 10^9 times 7 times 10^-24/6.40 times 10^-6 So, loss in kinetic energy = Gain in electrical potential energy 1/2 times 5.70 times 10^-14 times v^2 = 9 times 7 times 10^-9/6.40 or v^2 = 3.45 times 10^5 or v = 5.87 times 10^2 m/s