In the same example, what is the direction of the force on q3 if q1=2.0C , as in the example, but q2=3.0C ? Express your answer in degrees below the x axis.

SOLVE The net force acting on q3 is the vector sum of F1 and F2. We use Coulomb's law to find the magnitudes Fi and F2 of the forces: F1 = 91 93 T132 = (8.99 x 109 N . m2 /C2) (2.0x10 6 C) (4.0x10-6 C) (0.50 m)2 = 0.288 N F2 = 1 92 93 723 2 = (8.99 x 109 N . m2 /C2) (2.0x10 6 C)(4.0x10-6 C) (0.40 m)2 = 0.450 N We now calculate the x and y components of F1 and add them to the x and y components of F2, respectively, to obtain the components of the total force Ftotal on q3. From (Figure 1), sin 0 = (0.30 m) /(0.50 m) = 0.60 and cos 0 = (0.40 m) / (0.50 m) = 0.80. Since the y component of F2 is zero and its a component is positive, the a component of F2 is F2x = F2 = 0.450 N. The total a and y components are Ftotal,x = Fix + Fix = (0.288 N) cos 0 + 0.450 N = (0.288 N) (0.80) + 0.450 N = 0.680 N Ftotal,y = Fly + Fzy = -(0.288 N) sin 0 + 0 = -(0.288 N) (0.60) = -0.173 N These components combine to form Ftotal, as shown in (Figure 3): Ftotal = Ftotal, x + Foot = V(0.680 N)2 + (-0.173 N)2 = 0.70 N tan o Ftotal,y -0.173 N = Ftotal , 0.680 N and $ = -140 The resultant force has magnitude 0.70 N and is directed at 14 below the +x axis. REFLECT The forces exerted by q1 and q2 both have components in the +x direction, so these