Can Someone please help me out on this? I\'ve tried other ways that I\'ve found

Question

Can Someone please help me out on this? I've tried other ways that I've found but none of them are getting me the right answer! Much appreciated if someone could help out!! Thanks

Three charged particles are at the corners of an equilateral triangle as shown in the figure below. (Let q = 2.00 C, and L = 0.350 m.) 7.00 C 60.0 -4.00 C (a) Calculate the electric field at the position of charge q due to the 7.00-uc and-4.00- charges. kN/Cj (b) Use your answer to part (a) to determine the force on charge q mN j

Explanation / Answer

Q1 = 2 uC , Q2 = -4 uC , Q3 = 7 uC
Magnitude of force on Q1 due to Q3 = F = k*Q1*Q3/0.35² = 1.0286 N
Magnitude of force on Q1 due to Q2 = F = k*Q2*Q1/0.35² = - 0.588 N

Now you must add up the components in vector form:
Fx = Fcos(60°) - Fcos(60°) = - 0.588 *0.5 - 1.0286 *0.5 =-0.8083 N
Fy = Fsin(60°) + Fsin(60°) = - 0.588 *0.866 - 1.0286 *0.866 =-1.399 N
60° because it's an equilateral triangle.

F=E*Q1
E = F/Q1
E_x = Fx /Q1 = -404.15 kN/C ; E_y = Fy /Q1 = -699.5 kN/C (solution for part a )
b) force on Q1 :
    F = -808.3 mN i + -1399 mN j

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