Question

Can someone tell me where I'm going wrong? I have these two problems: The professor uploaded a video explaining how to solve them (generally, he didn't use numbers. He said the steps are the same for both problems except for the last step since the first problem is asking for magnitude and the second is asking for an angle. I followed his steps but I keep getting the wrong answer.) Also for the 2nd question I'm not sure if I'm drawing the picture correctly and if I should subtract from 360 or 180. Please see the 2nd picture for my work:

Question1 1) A charge, q1, of -9 micro-coulombs is located at the origin, a charge, q2, of -8 micro-coulombs is located atx -0 cm, y +6.2 cm, a charge, q3, of +9 micro- coulombs is located at xO cm, y-2.6 cm, and a charge, 44, of +4 micro-coulombs is located at x-6.3 cm, y 0 cm. What is the magnitude of the total electric force on q2 (charge on the ty axis) in Newtons? Account Dashboard Courses Correct Answer 64.55 margin of error +/-10% Calendar Inbox Question 2 0/18 pts Help A charge, q1, of-3 micro-coulombs is located at the origin, a charge, g2, of -6 micro coulombs is located at x 0 cm, y +5.7 cm, a charge, q3, of +4 micro-coulombs is located at x 0 cm, y6.5 cm, and a charge, q4, of +11 micro-coulombs is located atx 5.3 cm, y 0 cm. What is the angle of the total electric force on q2 (charge on the ty axis) in degrees measured counter-clockwise from the +x axis? Resources Correct Answer 208.6 margin of error+/-10% Question 3 0/18 pts A charge is located at the origin, anotr charge is located somewhere on the y

Answer 1

(1)

Electric force on q2 due to q1,

F1 = kq1q2 / r^2

F1 = 9*10⁹*9*10^-6*8*10^-6 / (0.062)²

F1 = 168.57j N (+y direction)

Electric force on q2 due to q3,

F3 = kq3q2 / r^2

F3 = 9*10⁹*9*10^-6*8*10^-6 / (0.088)²

F3 = 83.67 (-j) N (-y direction)

Electric force on q2 due to q4,

F3 = (kq2q4 / r^2)cos $\theta$ i +  (kq2q4 / r^2)sin $\theta$ -j

Here, $\theta$ = tan^-1 (6.2 / 6.3) = 44.54 deg

and r = sqrt (6.2² + 6.3²) = 8.839 cm

F3 = (9*10⁹*8*10^-6*4*10^-6 / (0.08839)²) * (cos 44.54 i - sin 44.54 j)

F3 = 26.273 i - 25.855 j

Net force on q2 in x direction,

Fx = 26.273 N

Net force on q2 in y direction,

Fy = 168.57 -  83.67 - 25.855 = 59.038 N

magnitude of net force on q2,

F = sqrt (26.273² + 59.038²)

F = 64.6 N

(2)

Electric force on q2 due to q1,

F1 = kq1q2 / r^2

F1 = 9*10⁹*3*10^-6*6*10^-6 / (0.057)²

F1 = 49.86j N (+y direction)

Electric force on q2 due to q3,

F3 = kq3q2 / r^2

F3 = 9*10⁹*4*10^-6*6*10^-6 / (0.122)²

F3 = 14.51 (-j) N (-y direction)

Electric force on q2 due to q4,

F3 = (kq2q4 / r^2)cos $\theta$ i +  (kq2q4 / r^2)sin $\theta$ -j

Here, $\theta$ = tan^-1 (5.7 / 5.3) = 47.082 deg

and r = sqrt (5.7² + 5.3²) = 7.78 cm

F3 = (9*10⁹*6*10^-6*11*10^-6 / (0.0778)²) * (cos 47.082 i - sin 47.082 j)

F3 = 66.76 i - 71.8 j

Net force on q2 in x direction,

Fx = 66.76 N

Net force on q2 in y direction,

Fy = 49.86 - 14.51 - 71.8 = -36.45 N

tan $\theta$ = Fy / Fx = 36.45 / 66.76

$\theta$ = 28.6 deg

$\theta$ = 28.6 + 180 = 208.6

$\theta$ = 208.6 deg counterclockwise from +x