Map Sapling Learning macmillan learning 18.4 uC 17.5 uC Pictured on the right ar

Question

Map Sapling Learning macmillan learning 18.4 uC 17.5 uC Pictured on the right are three point charges Q1 18.4 uC, Q2 =-38.6 JC, and Q3 = 57.3 h.C arranged according to the figure on the right. A fourth point charge is located at point A with a charge of QA = 17.5 Calculate the magnitude of the net force on the charge at point A. 30.1 cm Number 10.34 -38.6 I 30.1 cm >1 e Previous Try Again Next Exit Explanation Electric force is described by Coulomb's Law 1 4142 kq142 Here q1 and q2 are the two charges between which the force acts, r is the distance between the two charges, and Eo is the permittivity of free space. The Coulomb constant is k 1/(4TTEo). Force is a vector quantity, so the net force at a point in space is the sum of the force vectors. Force vectors can be added together by finding the x and y components each vector and summing the components in each direction. The final components can then be combined using the Pythagorean theorem to find the final magnitude

Explanation / Answer

force due to 18.4uC,

F1 = k Q1 / a^2 to the right

F1 = (9 x 10^9) (18.4 x 10^-6)(17.5 x 10^-6) / 0.301^2

F1 = 32 N  

due to Q3:

F3 = (9 x 10^9 x 57.3 x 10^-6 x 17.5 x 10^-6)/0.301^2

F3 = 99.6 N upward  

due to Q2:

F2 = (9x 10^9 x 38.6 x 10^-6 x 17.5 x 10^-6) / (0.301^2 + 0.301^2)

F2 = 33.55 N 45 deg below left side

Fx = F1 - F2 cos45 = 8.28 N  

Fy = F3 - F2 sin45 = 75.9 N  

F = sqrt[ Fx^2 + Fy^2] = 76.3 N .....Ans

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