Posted by **Beth** on Saturday, June 9, 2012 at 10:37pm.

Three charges are fixed to an xy coordinate system. A charge of +16C is on the y axis at y = +3.1 m. A charge of -16C is at the origin. Lastly, a charge of +55C is on the x axis at x = +3.1 m. Determine (a) the magnitude and (b) direction of the net electrostatic force on the charge at x = +3.1 m. Specify the direction as a positive angle relative to the +x axis.

I figured out how to set up the problem in general, but i keep getting the wrong answer. Please help!

So this is what I have gotten so far. I don't understand how to get the direction at all.

(9x10^-9)* (((18x10^-6)(55x10^-6))/ 3.1^2+3.1^2= 0.46357

- Physics 1 -
**Elena**, Sunday, June 10, 2012 at 4:21pm
q1 = 16 C, q2 =-16 C, q3 =55 C.

Distance between q1 and q3 r1= 3.1•√2 m.

Distance berween q2 and q3 r2 =3.1 m

Make the drawing: two forces acting on q3:

force F1 - along the “q1-q3” line - repulsion,

force F2 - along the “q2-q3” line - attraction.

F1 =k•q1•q3/r1² =9•10^9•16•55/(3.1•√2)²=4.12•10^11 N.

F2 =k•q2•q3/r1² =9•10^9•16•55/(3.1•2)²=8.24•10^11 N.

Net force is opposite the angle 45º according to the Cosine Law:

F12=sqrt(F1² +F2²-2•F1•F2•cos45 º) = 6.1•1011 N.

Two other angles in the “force triangle” we’ll find using Sine Law:

6.1•10^11/sin45=4.12•10^11/sinα,

sinα =0.478, α =28.5º

6.1•10^11/sin45=8.24•10^11/sinβ,

sinβ =0.94, β =70.2 º,

α+β =28.5+70.2=98.7 º - this is the angle that F12 makes with x-axis (below its positive direction)

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