Question

Two charges are located in the ?–? plane. If ?1=−3.20 nC and is located at (?=0.00 m,?=0.640 m), and the second charge has magnitude of ?2=3.80 nC and is located at (?=1.50 m,?=0.550 m), calculate the ? and ? components, ?? and ??, of the electric field, ?⃗ , in component form at the origin, (0,0). The Coulomb force constant is 1/(4??0)=8.99×109 N⋅m2/C2.

Answer 1

r₁ = distance of charge q₁ from origin = sqrt(x₁² + y₁²) = sqrt(0² + 0.640²) = 0.640 m

= angle of line joining the charge q₁'s location to the origin with x-axis = tan^-1(y₁/x₁) = tan^-1(0.640/0) = 90

r₂ = distance of charge q₂ from origin = sqrt(x₂² + y₂²) = sqrt(1.50² + 0.550²) = 1.6 m

= angle of line joining the charge q₂'s location to the origin with x-axis = tan^-1(y₂/x₂) = tan^-1(0.550/1.5) = 20.2 deg

E₁ = magnitude of electric field by charge q₁ = k q₁/r₁² = (9 x 10⁹) (3.20 x 10^-9)/(0.640)² = 70.3 N/C

E_1x = X-component of electric field by charge q₁ = E₁ Cos = (70.3) Cos90 = 0 N/C

E_1y = Y-component of electric field by charge q₁ = E₁ Cos = (70.3) Sin90 = 70.3 N/C

E₂ = magnitude of electric field by charge q₂ = k q₂/r₂² = (9 x 10⁹) (3.80 x 10^-9)/(1.6)² = 13.4 N/C

E_2x = X-component of electric field by charge q₂ = - E₂ Cos = - (70.3) Cos20.2 = - 65.98 N/C

E_2y = Y-component of electric field by charge q₂ = - E₂ Cos = - (70.3) Sin20.2 = - 24.3 N/C

Net electric field along x-direction is given as

E_x = E_1x + E_2x = 0 + (- 65.98) = - 65.98 N/C

Net electric field along y-direction is given as

E_y = E_1y + E_2y= 70.3 + (- 24.3) = 46 N/C