Question

The point charges in the figure below are located at the corners of an equilateral triangle 25 right.) cm on a side where% +24. and qe 3 95uc. Assume a the--a ls is directed to the (a) Find the electric field at the location of 9a magnitude direction ectric fields into their x and y components? N/C (counterclockwise from the +x-axis) Di Enter a number (b) What is the force on ar given that qa +1.50 nc? magnitude direction (counterclockwise from the +x-axis)

Answer 1

E_1 = q_c/4 pi elementof d^2 = 24 times 10^-6 times 9 times 10^9/(0.25)^2 = 3456 times 10^3 N/c E_2 = q_c/4 pi elementof d^2 = 3.95 times 10^-6 times 9 times 10^9/(0.25)^2 = 568.8 times 10^3 N/c E = (E_1 cos 60 degree +E_2 cos 60 degree)i + (E_1 sin 60 degree +E_2 sin 60 degree)j = (3456 times 10^3 cos 60 degree + 568.8 times 10^3 cos 60 degree)i + (3456 times 10^3 sin 60 degree - 568.8 times 10^3 sin 60 degree) E = 2012.4 times 10^3 i + 2500.4 times 10^3j Magnitude E = squareroot (2012.4 times 10^3)^2 + (2500.4 times 10^3)^2 = 3209.6 times 10^3 N/c Direction theta = tan^-1 [2500.4 times 10^3/2012.4 times 10^3] = 51.2 degree counter clockwise from + x-axis F = Eq_a = 3209.6 times 10^3 times 1.5 times 10^-9 F = 4.814 times 10^-3 N Direction theta = 51.2 degree counter clockwise from + -axis