Consider the following. (a) Red blood cells often become charged and can be trea
ID: 2002487 • Letter: C
Question
Consider the following.
(a) Red blood cells often become charged and can be treated as point charges. Healthy red blood cells are negatively charged, but unhealthy cells (due to the presence of a bacteria, for example) can become positively charged. In the figure, three red blood cells are oriented such that they are located on the corners of an equilateral triangle. The red blood cell charges are A = 2.30 pC, B = 7.10 pC, and C = 4.70 pC. Given these charges, what would the magnitude and direction of the electric field be at cell A? (1 pC = 1 1012 C.)
magnitude 2.25e5 N/C
direction° counterclockwise from the +x-axis
****PLEASE NOTE: I have everything except for the direction that is COUNTERCLOCKWISE from the x-axis. Please help me how to find the resultant direction. I have tried everything from tan-1(Ey/Ex), 360-tan-1(Ey/Ex), 180-tan-1(Ey-Ex), and 90-tan-1(Ey/Ex). Please show your work on how you get your answer. Thank you very much!
Explanation / Answer
electric field due to C = Ec = k*Qc/d^2 = 9*10^9*4.7*10^(-12)/(0.5*10^(-3))^2 = 16.92*10^4 N/C
and direction is east (toward +x axis)
so vector representation = Ec = 16.92*10^4 i
electric field due to B = Eb = k*Qb/d^2 = 9*10^9*7.1*10^(-12)/(0.5*10^(-3))^2 = 25.56*10^4 N/C
and direction is 30 deg west from -y axis
so vector representation = Eb = - 25.56*10^4*sin30 i - 25.56*10^4*cos30 j = - 12.78*10^4 i - 22.13*10^4 j
net electric field at A = E = Eb + Ec = - 12.78*10^4 i - 22.13*10^4 j + 16.92*10^4 i = 4.14*10^4 i - 22.13*10^4 j
magnitude of electric field = sqrt((4.14*10^4)^2 + (22.13*10^4)^2) = 22.51*10^4 = 2.25*10^5 N/C
direction from +x axis = arctan((-22.13*10^4)/(4.14*10^4)) = - 79.4 deg
direction counterclockwise from the x-axis = 360 - 79.4 = 280.6 deg
I think this is corret.