Question

Find the electric field magnitude and direction at the location of qo in the figure below, given that qb = 9.50 pC and qc--5.50 pC. Point charges located at the corners of an equilateral triangle 29.0 cm on a side. Ga gb Magnitude IE- direction (wrt +x-axis) =

Answer 1

E_b^vector = k q_b/r^2 = (9 times 10^9) (9.5 times 10^-6)/0.29^2 = 1.0166 times 10^6 N/c E_c^vector = (9 times 10^9) (5.50 times 10^-6)/(0.29)^2 = 0.5886 times 10^6 N/c E_net^vector x = (1.0166 times 10^6 times cos 60 degrees) + (0.5886 times 10^6 times cos 60 degrees) = -0.8026 times 10^6 N/c E_net^vector y = 1.0166 times 10^6 times sin 60 degrees - 0.5886 times 10^6 times sin 60 degrees = 0.370676 times 10^6 N/c E_net = squareroot (E_x)^2 + (E_y)^2 = 8.84 times 10^5 N/c theta = 180 degrees - tan 0.8026/0.370676 = 114.8 degrees with +ve F_x^vector = 0.8026 times 10^6 times 1.6 times 10^-9 = -1.284 times 10^-3 N F_y^vector = 0.370676 times 10^6 times 1.6 times 10^-9 = 0.5931 times 10^-3 N F_net^vector = squareroot F_x^2 + F_y^2 = 1.414 times 10^-3 N theta = 180 degrees - tan 0.5931/1.284 = 155.207 degrees