Question

PRINTER VERSTONBACK Chapter 18, Problem 34 GO Your answer is partially correct. Try again. The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 7.20 cm from the origin O. For each of the situations in the drawing, determine the magnitude of the net electric field at the origin. -5.0 AC 1.0rC +2.0prC 3.0C 4.0 -1.0AC 0 6.0rC 0 (a)E1.2E7 N/C N/C (b)12E7 Click if you would like to Show Work for this question: Question Attempts: 1 of 5 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

a)

1.0p 2 4.0AC Ks 20C 2, 6.0p (0

r = distance of each charge from origin = 7.20 cm = 0.072 m

In case 1

E₁ = electric field by charge Q₁ = k Q₁/r² = (9 x 10⁹) (2 x 10^-6)/(0.072)² = 3.5 x 10⁶ N/C

E₂ = electric field by charge Q₂ = k Q₂/r² = (9 x 10⁹) (3 x 10^-6)/(0.072)² = 5.21 x 10⁶ N/C

E₃ = electric field by charge Q₃ = k Q₃/r² = (9 x 10⁹) (5 x 10^-6)/(0.072)² = 8.68 x 10⁶ N/C

Net electric field along x-direction is given as

E_x = E₁ + E₂ = (3.5 x 10⁶) + (5.21 x 10⁶) = 8.71 x 10⁶ N/C

Net electric field along y-direction is given as

E_y = E₃ = 8.68 x 10⁶ N/C

Net electric field is given as

E = sqrt(E_x² + E_y²) = sqrt((8.71 x 10⁶)² + (8.68 x 10⁶ )²) = 12.3 x 10⁶ N/C

b)

E₁ = electric field by charge Q₁ = k Q₁/r² = (9 x 10⁹) (4 x 10^-6)/(0.072)² = 6.94 x 10⁶ N/C

E₂ = electric field by charge Q₂ = k Q₂/r² = (9 x 10⁹) (1 x 10^-6)/(0.072)² = 1.74 x 10⁶ N/C

E₃ = electric field by charge Q₃ = k Q₃/r² = (9 x 10⁹) (1 x 10^-6)/(0.072)² = 1.74 x 10⁶ N/C

E₄ = electric field by charge Q₄ = k Q₄/r² = (9 x 10⁹) (6 x 10^-6)/(0.072)² = 10.4 x 10⁶ N/C

Net electric field along x-direction is given as

E_x = E₁ + E₂ = (6.94 x 10⁶) + (1.74 x 10⁶) = 8.68 x 10⁶ N/C

Net electric field along y-direction is given as

E_y = E₄ - E₃ = 10.4 x 10⁶ - 1.74 x 10⁶ = 8.66 x 10⁶ N/C

Net electric field is given as

E = sqrt(E_x² + E_y²) = sqrt((8.68 x 10⁶)² + (8.66 x 10⁶ )²) = 12.3 x 10⁶ N/C