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Three point charges are arranged as shown in the figure below. (Take q1 = 5.58 nC, q2 = 4.60 nC, and q3 = -2.82 nC.)

(a) Find the magnitude of the electric force on the particle at the origin.
N

(b) Find the direction of the electric force on the particle at the origin.
degrees (counterclockwise from the +x-axis)
Edit
Added Fri, 03 Jul '15
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1.(a) the resultant force would be found by vector addition of aal the forces acting at origin.

so, in this case there are only 2 forces ...

F1 force due to q1 will act alonf -x axis (repulsive nature)

F2. force due to q3 wiil act along -ve y axis (attaractive nature).

F1=k*q1*q2/r*r= 4.6*10-9 *5.58*10-9 *9*109 /0.3*0.3=2.556*10-6 N(-i)

F2=9*109 * 4.6*10-9 *2.82*10-9/0.1*0.1=11.6*10-6 N(-j)

so net force is F=2.556*10-6(-i) +11.6*10-6(-j) N

magnitude is F=((F1)2+(F2)2)1/2 =11.87*10-6 N

1(b)

the angle that resultant force willl make with +x axis is tan-1(11.6/2.556)=77.57 degree

but the resultant is in third quadrant so total angle that wiil be made along counter clockwise dirn wiil be 180+77.57=257.57 degrees
Edit
Added Fri, 03 Jul '15
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