Question

The figure shows two charged particles on an x axis: -q = -6.40 × 10-19 C at x = -2.00 m and q = 6.40 × 10-19 C at x = 2.00 m. What are the (a)x- and (b)y- components of the net electric field produced at point P at y = 3.00 m?

Answer 1

Given

Q = 6.4*10^-19 C

------------------------------------------------

From the diagram

AO = 2.00 m, OC = 3.00 m

theta = tan^-1(OC/AO) = 63.43^0

AC = sqrt(AO^2 + OC^2) = 3.6 m

-----------------------------------------------

The electric field due to -Q at point C

E_1 = (1/4*pi*e_0)*Q/(AC)^2

E_1 = 9*10^9 * 6.4*10^-19 / (3.6)^2

E_ 1 = 4.44*10^-10 N/C

E_1x = E_1cos243.43^0 = -1.986 *10^-10N/C

E_1y = E_1sin243.43^0 = -3.97*106-10 N/C

The electric field due to -Q at point C

E_2 = (1/4*pi*e_0)*Q/(BC)^2

E_2 = 9*10^9 * 6.4*10^-19 / (3.6)^2

E_ 2 = 4.44*10^-10 N/C

E_1x = E_1cos153.43^0 = -3.97 *10^-10N/C

E_1y = E_1sin153.43^0 = 1.986*106-10 N/C

The horizontal components of net field

E_x = E_1x + E_2x

The verticle components of net field

E_y = E_1y + E_2y

Plug the values and get the answer