Question

Consider the following. 0.500 mm 60.00 (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 = 2.40 PC, B = 6.90 pc, and C = -4.80 PC. Given these charges, what would the magnitude and direction of the electric field be at cell A? (1 PC = 1 x 10-12 C.) magnitude N/C direction counterclockwise from the +x-axis (b) If the charge of cell A were doubled, how would the electric field at cell A change? The magnitude of the field would be quadrupled. The magnitude of the field would be halved. The magnitude of the field would be doubled. The field would be unchanged.

Answer 1

Prob. у Pol 0.5 mm A 600 » X Ecla U EBIA PA = 2.4 PC = 2.4 X 10-126 AB=BC=CA=0.5 mm. 5x109 23= 6.9 PC → 6.941012 ac- 4.8x10-12c. 4.8pc Net electric field at A vector sum of EB L EC due to charges at B &c. due to charge B. 98 9x109 x EB I 4ME 82 6.98 10 12 (5810-4)2 EBE - 248400 N/ due to change ci ac Ec= ㅗ 4 TEO 72

Ec re ㅗ 40C 72 9x109 x 4.8x1012 (1024x5)2 Ec= 192800 N/c P-2 SO А Ec ud 120° EB EA Resultant EA = (68)+(Ec)?+ 2GB! E COSI 20° LA 4.8638 x lolo EA = 220542:24 Nic Direction EB tand = 55:2 Ec counter clock wise 360-d 30 4.82 °
The net electric field at point A will be the vector sum of the electric field due to point charge B and point charge C which we can find by applying the coulamb formula.

(b). If chage at A is doubled there will be no change in thr electric field at point A because the electric field at A is due to other 2 charges B and C. Charge A does not effect the electric field at point A.

so correct option is 'd'.

Please find the attached solution and provide the feedback.