Three point charges, A = 1.75 mu C, B = 6.80 mu C, and C = -4.10 mu C, are locat

Question

Three point charges, A = 1.75 mu C, B = 6.80 mu C, and C = -4.10 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 1.75 mu C charge. magnitude N/C direction degree below the +x-axis (b) How would the electric field at that point be affected if the charge there were doubled? The magnitude of the field would be halved. The field would be unchanged. The magnitude of the field would double. The magnitude of the field would quadruple. Would the magnitude of the electric force be affected? Yes No

Explanation / Answer

(a)

x component of Electric field due to B = EBx = -k*qB*cos60/l^2 = -(9*10^9*6.8*10^-6*cos60)/0.5^2 = -122400 N/C

y component of Electric field due to B = EBy = -k*qB*sin60/l^2 = -(9*10^9*6.8*10^-6*sin60)/0.5^2 = -212003 N/C

x component of Electric field due to C = EBx = k*qc/l^2 = (9*10^9*4.1*10^-6)/0.5^2 = 147600 N/C

y component of Electric field due to C = EBy = 0

Ex = Ebx + Ecx = 25200 N/C

Ey = Eby + Ecy = -212003 N/C

(a)

magnitude E = sqrt(Ex^2 + Ey^2 ) = 213495.5 N/C = 2.13*10^5 N/C

direction = tan^-1(Ey/Ex) = 83.22

(b)

if charges are doubled

E'x = 2Ex

E'y = 2Ey

magnitde = 2*E

The magnitude of the field would double

(c)Yes

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