The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 2.37 µC, x1 = 5.87 cm, y1 = 0.494 cm and92 = -6.49 µC, x2 = -2.14 cm, y2 = 1.12 cm. Find the (a) magnitude and (b) direction (with respect to +x-axis in the range(-180°;180°) of the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle ofcharge q3 = 6.63 µC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?(a) NumberiUnits(b) NumberiUnits(c) NumberiUnits(d) NumberiUnits>> •15 Go The charges and coordinates of two charged particles heldfixed in an xy plane are q = +3.0 µC, x1 = 3.5 cm, y = 0.50 cm,and q2 = -4.0 µC, x, = -2.0 cm, y, = 1.5 cm. Find the (a) magni-tude and (b) direction of the electrostatic force on particle 2 due toparticle 1. At what (c) x and (d) y coordinates should a third parti-cle of charge q3 = +4.0 µC be placed such that the net electrostaticforce on particle 2 due to particles 1 and 3 is zero?Your answer is partially correct.The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 3.11 µC, x1 = 5.01 cm, y1 = 0.163 cm and92 = -3.84 µC, x2 = -2.24 cm, y2 = 1.95 cm. Find the (a) magnitude and (b) direction (with respect to +x-axis in the range (-180°;180°]) ofthe electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge q3 - 4.86 µCbe placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?(a) Number19.3UnitsN15. (a) The distance between qi and q2 is(b) Number-13.8Units* (degrees)rz =(x, -x,)* +(y, -x) =(-0.020 m–0.035 m)* +(0.015 m–0.005 m)* =0.056 m.The magnitude of the force exerted by q, on q; is(c) NumberUnits-0.095m(8.99×10°N-m²/C) (3.0×10 C) (4.0×10ʻC)(0.056 m)F =k=35 N.(d) Number0.037Unitsm(b) The vector Ē, is directed toward gj and makes an angle Owith the +x axis, where972СНАРТER 211.5 cm-0.5 cm0= tan -= tan=-10.3°-10°.-2.0 cm-3.5 cm(c) Let the third charge be located at (x3, ya), a distance r from q2. We note that qı, 92, andq3 must be collinear; otherwise, an equilibrium position for any one of them would beimpossible to find. Furthermore, we cannot place q3 on the same side of q2 where we alsofind q1, since in that region both forces (exerted on q2 by q3 and qi) would be in the samedirection (since qz is attracted to both of them). Thus, in terms of the angle found in part(a), we have x =x -r coso and y "y2 -r sino(which means y3 > y2 since O is negative).The magnitude of force exerted on q2 by q3 is F=k|q:9, Vr, which must equal that ofthe force exerted on it by q, (found in part (a)). Therefore,= 0.0645m = 6.45 cmConsequently, x3 =x2 -r cos0=-2.0 cm - (6.45 cm) cos(-10°) =-8.4 cm,(d) and y; = y2 -r sino= 1.5 cm – (6.45 cm) sin(-10°) = 2.7 cm.

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