Question

Two charges are placed on the x axis. One of the charges (q₁ = +8.4 µC) is at x₁ = +2.7 cm and the other (q₂ = -24 µC) is at x₂ = +8.4 cm.​

Find the net electric field (magnitude and direction) at x = +6.4 cm. (Use the sign of your answer to indicate the direction along the x-axis.)​

Answer 1

An electric field due to charge q₁ which will be given as -

E₁ = k_e |q₁| / (x - x₁)²

E₁ = (9 x 10⁹ Nm²/C²) (8.4 x 10^-6 C) / [(0.064 m) - (0.027 m)]²

E₁ = [(75600 Nm²/C) / (0.001369 m²)]

E₁ = 5.52 x 10⁷ N/C

An electric field due to charge q₂ which will be given as -

E₂ = k_e |q₂| / (x - x₂)²

E₂ = (9 x 10⁹ Nm²/C²) (24 x 10^-6 C) / [(0.064 m) - (0.084 m)]²

E₂ = [(216000 Nm²/C) / (0.0004 m²)]

E₂ = 54.0 x 10⁷ N/C

Magnitude of the net electric field at a distance which will be given as :

E_net = E₁ + E₂

E_net = [(5.52 x 10⁷ N/C) + (54.0 x 10⁷ N/C)]

E_net = 5.95 x 10⁸ N/C

(Direction along the +x axis)