Question

A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x = 0.60 m , y = 0.

A) Calculate the magnitude of the net electric field at the origin due to these two point charges

B) Calculate the direction of the net electric field at the origin due to these two point charges.

Answer 1

A)

In triangle ABC

r₁ = distance of charge q₁ from origin = sqrt(0.60² + 0.80²) = 1 m

= tan^-1(AC/BC) = tan^-1(0.80/0.60) = 53.12 deg

r₂ = distance of charge q₂ from origin = 0.60 m

magnitude of electric field by charge q₁ is given as

E₁ = k q₁/r₁² = (9 x 10⁹) (4 x 10^-9)/(1)² = 36 N/C

magnitude of electric field by charge q₂ is given as

E₂ = k q₂/r₂² = (9 x 10⁹) (6 x 10^-9)/(0.60)² = 150 N/C

Net electric field along x-direction is given as

E_x = E₁ Cos - E₂ =  (36) Cos53.12 - 150 = - 128.4 N/C

Net electric field along y-direction is given as

E_y = E₁ Sin=  (36) Sin53.12 = 28.8 N/C

Net electric field is given as

E = sqrt(E_x² + E_y²) = sqrt((- 128.4)² + (28.8)²) = 131.6 N/C

B)

= direction = 180 - tan^-1(E_y/E_x) = 180 - tan^-1(28.8/128.4) = 167.4 deg counterclockwise from +X-direction