Three point charges, A = 1.95 muC, B = 7.15 muC, and C = -4.55 muC, are located

Question

Three point charges, A = 1.95 muC, B = 7.15 muC, and C = -4.55 muC, 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 1.95 pC charge. Magnitude N/C direction degree below the +x-axis 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

(a) k =9*10^9 N.m^2/C^2

qA = 1.95 uC, qB =7.15 uC,qC = -4.55 uC,d = 0.5 m

F = kq^2/r^2

Attraction force between A and C is

FAC = (9*10^9*1.95*4.55*10^-12)/(0.5*0.5)

FAC = 0.31941 i

Repulsive force between A and B is

FAB = (9*10^9*1.95*7.15*10^-12)/(0.5*0.5) = 0.50913

FAB = - 0.50913 sin(60) i- 0.50913 cos(60) j

FAB = -0.441 i - 0.255 j

Net force on A is F = FAB +FAC

F = -0.441 i - 0.255 j + 0.31941 i

F = -0.122 i -0.255 j

Magnitude of force = [(-0.122)^2+(-0.255)^2]^1/2

Magnitude of force F = 0.283 N

direction tan(theta) = 0.255/0.122

theta = 64.43

direction = 64.43+180 =244.43 degrees with + x axis

(b) The filed would be unchanged

(c) NO

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