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W20 of t DMSN-0: Mail a Watch Onli | Watch Blac Are you | G rachel c lon capa.auamed.net/enc/50/1ffc884 16d67632a3d995 Q1 (0,0) Q2 As shown above, Q1 is a charge of-8.9pC, located at x=0, y 2.3 cm. Q2 is a charge of +2.7uC, located at x-0, y- -5.1 cm. Point A is at a location of x-14 cm, y=0. Calculate the magnitude of the electric field due to Q1 at point A. Submit Answer Tries 0/10 Calculate the angle of the electric field due to Q1 at point A relative t (Use deg as the units) Submit Answer Tries 0/10 Calculate the magnitude of the electric field due to Q2 at point A. Submit Answer Tries o/10 Calculate the angle of the electric field due to Q2 at point A relative to the +x axis. Submit Answer Tries 0/10 Calculate the magnitude of the total electric field at point A. Submit Answer Tries 0/10 Calculate the angle of the total electric field at point A relative to the +x axis.

Explanation / Answer

r1 = distance betwen Q1 and point A = sqrt((14-0)^2+(0-2.3)^2) = 14.2 cm = 0.142 m

r2 = distance betwen Q2 and point A = sqrt((14-0)^2+(0+5.1)^2) = 14.9 cm = 0.149 m

magnitude of electric field due to Q1 at point A

E1 = k*Q1/r1^2 = 9*10^9*8.9*10^-6/0.142^2 = 3.97*10^6 N/C

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angle of the electric field due to Q1 at point A relative to +x axis

tantheta1 = (2.3/-14)

theta1 = 170.67

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magnitude of electric field due to Q2 at point A

E2 = k*Q2/r2^2 = 9*10^9*2.7*10^-6/0.149^2 = 1.1*10^6 N/C

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angle of the electric field due to Q2 at point A relative to +x axis

tantheta2 = (5.1/14)

theta2 = 20

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x component of total field Ex = E1x + E2x

Ex = E1*costheta1 + E2*costheta2

Ex = (3.97*10^6*cos170.67)+(1.1*10^6*cos20)

Ex = -2.88*10^6 N/C

y component of total field

Ey = E1y + E2y

Ey = E1*sintheta1 + E2*sintheta2

Ey = (3.97*10^6*sin170.67)+(1.1*10^6*sin20)

Ey = 1.02*10^6 N/C

magnitude of total electric field E = sqrt(Ex^2+Ey^2)

E = 3.05*10^6 N/C

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angle of total electric field

tantheta = Ey/Ex

theta = tan^-1(Ey/Ex) = 160.5 degrees

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