Question

A 2.14 10-9 C charge has coordinates x = 0, y = −2.00; a 3.09 10-9 C charge has coordinates x = 3.00, y = 0; and a -5.25 10-9 C charge has coordinates x = 3.00, y = 4.00, where all distances are in cm. Determine magnitude and direction for the electric field at the origin and the instantaneous acceleration of a proton placed at the origin. (a) Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis). magnitude Incorrect: Your answer is incorrect. direction Incorrect: Your answer is incorrect. ° (b) Determine the magnitude and direction for the instantaneous acceleration of a proton placed at the origin (measure the angle counterclockwise from the positive x-axis). magnitude direction °

Answer 1

E₁ = k q₁/(OC)² = (9 x 10⁹) (2.14 x 10^-9) / (0.02)² = - 48150 N/C

E₂ = k q₂/(OB)² = (9 x 10⁹) (3.09 x 10^-9) / (0.03)² = - 30900 N/C

E₃ = k q₃/(OA)² = (9 x 10⁹) (5.25 x 10^-9) / (0.05)² = 18900 N/C

E_3x = X-component of electric field by charge q₃ = 18900 Cos53 = 11374.3 N/C

E_3y = Y-component of electric field by charge q₃ = 18900 Sin53 = 15094.2 N/C

Net electric field along X-axis

E_x = E₂ + E_3x = - 30900 + 11374.3 = - 19525.7 N/C

Net electric field along Y-axis

E_y = E₁ + E_3y = - 48150 + 15094.2 = - 33055.8 N/C

Net electric field

E = sqrt (E_x² + E_y²) = sqrt ((- 19525.7 )² + (- 33055.8 )²) = 38392 N/C

direction = = tan^-1(E_y/E_x) = tan^-1(-33055.8/-19525.7) = 59.5 fron negative x-axis

180 + 59.5 from positive X-axis in countercockwise direction

b)

F = q E = (1.6 x 10^-19) (38392 ) = 6.14 x 10^-15

direction : 180 + 59.5 from positive X-axis in countercockwise direction