Shows three point charges, Q_1 = 3.0 muC, Q_2 = -5.0 muC, Q_3 = 8.0 muC located

Question

Shows three point charges, Q_1 = 3.0 muC, Q_2 = -5.0 muC, Q_3 = 8.0 muC located at the corners of an equilateral triangle of side length 0.50 m. Q_1 is at the origin, Q_2 is on the x-axis at x = 0.5 m & Q_3 is at the top vertex of the triangle as shown. As the position of the charge Q_3 = 8.0 muC, calculate (numerically): a. The x-component E_x1 of the electric field due to Q_1 and the x-component E_x2 of the electric field due to Q_2. b. The y-component E_y1 of the electric field due to Q_1 and the y-component E_y2 of the electric field due to Q_2. c. The x-component E_x and the y-component E_y of the total electric field due to Q_1 and Q_2. d. The magnitude and direction of the total electric field due to Q_1 and Q_2. e. The total electric potential due to Q_1 and Q_2.

Explanation / Answer

a. magnitude of electric field due to Q1 = kQ1/a^2

E1 = (9 x 10^9 x 3 x 10^-6)/0.5^2 = 108000 N/C

at angle 60 deg above x axis.

E1x = E1cos60= 54000 N/C

E1y = E1sin60 = 93530.7 N/C

b. E2 = (9 x 10^9 x 5 x 10^-6)/0.5^2 = 180000 N/C

at angle 60 deg below x axis.

E2x = E1cos60= 90000 N/C

E2y = - E1sin60 = - 155884.5 N/C

c.
Ex = E1x + E2x = 144000 N/C

Ey = - 62353.8 N/C

d. magnitdue = sqrt(Ex^2 + Ey^2) = 156920.35 N/C

direction = tan^-1(Ey / Ex) = - 23.41 deg (below +ve x axis)

e. V = kQ1/a + kQ2/a

   = ( 9 x 10^9) ( 3-5 x 10^-6) / 0.5

   = -36000 Volt

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