Taking the (positive) square root and solving, we obtain x= 0.683 m. If one takes the negative root and ‘solves’, one finds the location where the net force wouldbe zero ifq1and q2were of like sign (which is not the case here). (d) From the above, we see that y= 0. 64. Charge q1= –80 ×10–6C is at the origin, and charge q2= +40 ×10–6C is at x= 0.20 m. The force on q3= +20 ×10–6C is due to the attractive and repulsive forces from q1and q2, respectively. In symbols, GGGFFF33132net=+, where 313231322232||,| |.qqqqFkFkrr==GG(a) In this case r31= 0.40 m and r32= 0.20 m, with FGdirected towards –xand FGdirected in the +xdirection. Using the value of kin Eq. 21-5, we obtain 32123net322669226ˆˆˆˆi| |iii80 10 C40 10 Cˆ(8.99 10 N m C )(20 10 C)i(0.40m)(0.20m)ˆ(89.9 N)i .qqFkkkqrr−−−⎛⎞⎛⎞=−+= −+=−+⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠−×+×=×⋅×+=G(b) In this case r31= 0.80 m and r32= 0.60 m, with 31FGdirected towards –
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