Question

The figure shows two charged particles on an x axis: -q = -6.40 × 10^-19 C at x = -4.50 m and q = 6.40 × 10^-19 C at x = 4.50 m. What is the x-component of the net electric field produced at point P at y = 4.00 m?

Answer 1

The figure shows that two charged particles on an x axis is given below as :

E+ E- 0 -9

Distance, r² = (4.5 m)² + (4 m)² = 36.2 m²

The x-component of the net electric field produced at point P which will be given as :

E_net,x = E₊ + E_-

E_net,x = 2 (k_e q / r²) cos $\theta$                                                      (SINCE, cos $\theta$ = x / r)

where, q = charge on particle = 6.4 x 10^-19 C

then, we get

E_net,x = 2 [(9 x 10⁹ Nm²/C²) (6.4 x 10^-19 C) / (36.2 m²)] [(4.5 m) / (6.01 m)]

E_net,x = (3.18 x 10^-10 N/C) (0.7487)

E_net,x = 2.38 x 10^-10 N/C