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You have 2 days left in your Cengage Unlimited subscription trial.Leam more for less with access to over 22,000 ebonks and digtal learning product Up 4.. 0/2 points 1 s Answers SerPSET9 23.P015.MI Three charged particles are located at the corners of an equilateral triangle as shown in 0.670 m). Calculate the total electric force on the 7.00-uC charge. the figure below (let q-2.20 yC, and L 0.266 If you calculate the magnitude of the force that each charge exerts on the 7.00 uC charge, the net charge can then be found from the vector sum of those forces. N 85.27 direction You need the components of the total force in order to find this angle. (counterclockwise from the +x axis) 7.00 C 60.0 4.00 C Need Help? ReadItMaster t Read It My Notes .

Explanation / Answer

electrostatic force is given by

F = kq1*q2/d^2

Force will be attractive if both charge have different signs, and force will be repulsive if both charge have same signs. Now Net force will be

qa = 7 uC, qb = 2.20 uC & qc = -4.00 uC

rab = rbc = L = 0.670 m, So

Fnet = Fab + Fac

Fab = Force between qa and qb, which will be repulsive at 60 deg above +x-axis

Fac = Force between qa and qc, which will be attractive at 60 deg below +x-axis

Fab = k*qa*qb/L^2

Fab = 9*10^9*7*10^-6*2.20*10^-6/0.670^2

Fab = 0.309 N

Fabx = Fab*cos 60 deg = 0.309*cos 60 deg = 0.1545 N

Faby = Fab*sin 60 deg =  0.309*sin 60 deg = 0.2676 N

Now

Fac = k*qa*qc/L^2

Fac = 9*10^9*7*10^-6*4.00*10^-6/0.670^2

Fac = 0.561 N

Facx = Fac*cos 60 deg = 0.561*cos 60 deg = 0.2805 N

Facy = -Fac*sin 60 deg = -0.561*sin 60 deg = -0.4858 N

Now Net force will be

Fnet = Fab + Fac

Fnet = (Fabx + Facx) i + (Faby + Facy) j

F = (0.1545 + 0.2805) i + (0.2676 - 0.4858) j

F = 0.435 i - 0.2182 j

Magnitude of Force will be

|F| = sqrt (0.435^2 + (-0.2182)^2)

|F| = 0.487 N

direction = arctan (0.2182/0.435) = 26.64 deg below positive x-axis

direction = 360 - 26.64 = 333.36 deg CCW from +x-axis

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