1. A charge of +3.5 C is placed on the x axis at x = +0.55 m while a charge of –

1. A charge of +3.5 C is placed on the x axis at x = +0.55 m while a charge of -15 C is placed at the origin. (a) Calculate the magnitude and direction of the net electric field on the x-axis at x = +0.8 m. (b) Determine the magnitude and direction of the force that would act on a charge of 8.0 C if it was placed on the x axis at x = + 0.8 m. Solution:a) The electric field from a point charge is given by E = k|q|/r^2 E points towards a charge and array from a + charge. Both electric field points in the same direction so we just add them together E (net) =(8.99*10^9 Nm^2/C^2 * 3.5*10^-6 C)/ 0.55^2 + (8.99*10^9 Nm^2/C^2 * 15*10^-6 C)/ 0.8^2 E (net) = 2.9* 10^5 N/C +x directionb) ?