Question

(35pts.) Point charges qu = +4uC, q2 = -391, and 93 = +2qı are placed as shown in the figure below with the distance a = 6cm. These charges. together with point P, form an isosceles triangle as shown in the figure below. Find the net electric field at point P.

Answer 1

here,

q1 = 4 uC = 4 * 10^-6 C

q2 = - 3 q1 = - 12 uC = - 12 * 10^-6 C

q3= 2 q1 = 8 uC = 8 * 10^-6 C

a = 6 cm = 0.06 m

theta = arctan(a/a) = 45 degree

the electric feild due to q1 , E1 = K * q1 /a^2 j

E1 = 9 * 10^9 * 4 * 10^-6 /0.06^2 j N/C

E1 = 10 j N/C

the electric feild due to q2 , E2 = K * q2 /(a^2 + a^2) * ( - cos(45) i - sin(45) j) j

E2 = 9 * 10^9 * 12 * 10^-6 /(0.06^2 * 1.41 ) * 0.71 * ( - i - j) N/C

E2 = (- 15 i - 15 j) N/C

the electric feild due to q3 , E3 = K * q3 /(a^2 + a^2) * ( - cos(45) i + sin(45) j) j

E3 = 9 * 10^9 * 8 * 10^-6 /(0.06^2 * 1.41 ) * 0.71 * ( - i + j) N/C

E3 = (- 10 i - 10 j) N/C

the net electric feild , E = E1 + E2 + E3

E = ( -25 i - 15 j) N/C

the magnitude of net electric feild , |E| = sqrt(25^2 + 15^2) = 29.2 N/C

the direction of electric feild ,theta = arctan((-15)/(-25)) = 211 degree counterclockwise from +X axis