Three point charges, A = 1.80 mu C, B = 7.40 mu C, and C = - 4.20 mu C, are loca

Question

Three point charges, A = 1.80 mu C, B = 7.40 mu C, and C = - 4.20 mu C, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 1.80 mu C charge. How would the electric field at that point be affected if the charge there were doubled? The magnitude of the field would be halved. The field would be unchanged. The magnitude of the field would double. The magnitude of the field would quadruple. Would the magnitude of the electric force be affected? Yes No

Explanation / Answer

FBA is the force of B on A

FBA = (9 x 10^9 x 1.8 x 10^-6 x 7.4 x 10^-6) / 0.5^2 = 0.4795 N (direction away from B)

FCA is the force of C on A. this lies on the x-axis

FCA = (9 x 10^9 x 1.8 x 10^-6 x -4.2 x 10^-6) / 0.5^2 = -0.2722 N (direction towards C)

x-component of FBA = 0.4795cos(60) = -0.2398N (-x direction)

y-component of FBA = 0.4795sin(60) = -0.4153N (-y direction)

x-component of FCA = 0.2722N (+x direction)

y-component of FCA = 0

Total x component = -0.2398 + 0.2722 = 0.0324N

Total y component = -0.4153N

Resultant = sqrt(0.0324^2 + 0.4153^2)

= 0.4166N

Angle to the x axis = tan^-1(-0.4153/0.0324) = -85.54degrees (= 274.46degrees)

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