Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

please help thank you Consider the following. (a) Red blood cells often become c

ID: 1375804 • Letter: P

Question

please help thank you 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 = 2.10 pC, B = 6.70 pC,

and

C = ?4.00 pC.

Given these charges, what would the magnitude and direction of the electric field be at cell A? (1 pC = 1 ? 10?12 C.)


(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 halved.The magnitude of the field would be quadrupled.    The magnitude of the field would be doubled.The field would be unchanged.

magnitude     N/C direction     ? counterclockwise from the +x-axis please help thank you 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 = 2.10 pC, B = 6.70 pC, and C = ?4.00 pC. Given these charges, what would the magnitude and direction of the electric field be at cell A? (1 pC = 1 ? 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 halved.The magnitude of the field would be quadrupled. The magnitude of the field would be doubled.The field would be unchanged.

Explanation / Answer

i see part a is already solved

the answer for part b is

the field would remain unchanged

this is because the field at A is due to the other two charges