Red blood cells often become charged and can be treated as point charges. Health

Question

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 = 2.40 pC, B = 6.90 pC and C = ?4.20 pC.

Given these charges, what would the magnitude of the electric field be at cell A?

I get 2.499e+05 on magnitude but not correct.

Explanation / Answer

Elecric filed at A due to charge at B is E1 = k*qB/r1^2

giVen that qB = 6.9pC
r1 = 0.5*10^-3 m and k = 9*10^9

E1 = (9*10^9*6.9*10^-12)/(0.5*10^-3)^2 = 248400 N/C

Electric field at A due to charge at C is E2 = k*qC/r2^2

giVen that qC = 4.2pC
r2 = 0.5*10^-3 m and k = 9*10^9

E2 = (9*10^9*4.2*10^-12)/(0.5*10^-3)^2 = 151200 N/C

net Electric field is E= sqrt[E1^2+E2^2+(2*E1*E2*cos(60))]

E = sqrt(248400^2+151200^2+(2*248400*151200*cos(60))) = 349459.7 N/C

E = 3.49*10^5 N/C

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