Three point charges, A = 1.70 mu C, B = 6.70 mu C, and C = -4.30 mu C, are locat

Question

Three point charges, A = 1.70 mu C, B = 6.70 mu C, and C = -4.30 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 1.70 mu C charge. your response differs from the correct answer by more than 10%. Double check your calculations. N/C Your response differs from the correct answer by more than 10%. Double check your calculations. degree (counterclockwise from the +x-axis) How would the electric field that point be affected if the charge there were doubled?

Explanation / Answer

Given

   charges at the three corners of an equilateral triangle of side r = 0.5 m

qA = 1.7*10^-6 C, qB = 6.7*10^-6 C , qC = -4.30*10^-6 C

magnitude and direction of E at the position where 1.7*10^-6 C situated

we know that the electric field due a charge q at a distance r is E = kq/r^2

now E at A due to qB is E1 = kqB/r^2 = 9*10^9*6.7*10^-6/(0.5)^2 N/C = 241200 N/C

the direction is along the line joining the charges qA,qB which is at an angle of 240 degrees with x axis

so E1x = E1 cos theta
   =241200*cos 240 = -120600 N/c

E1y = E1 sin theta = 241200*sin 240 = -208885.3274 N/C

and the field due to chargeqC at A is

   E2 = kqC/r^2 = 9*10^9*4.3*10^-6/(0.5)^2 N/C = 154800N/C

the direction is along +x axis

so    E2x = E2 cos 0 = 154800 N/C ,
   E2y = E2 sin 0 = 0 N/C

net field is E = Ex+Ey

       = (-120600+154800)i+(-208885.3274)j
magnitude is

   E = sqrt(Ex^2+Ey^2)
   E = sqrt((34200)^2+(-208885.3274)^2) N/C =211666.530 N/C

the direction is

   theta = arc tan((-208885.3274)/(34200))

   =-80.70 degrees below the x axis

or 280 degrees from the x axis

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