Find Distance of Electric Charge 3 from q1: 13.9 cm

[tony873004]

3. Suppose the charge q2 in Figure 19-30 can be moved left or right along the line connecting the charges q1 and q3, rather than sitting at a distance d from each. Given that q = 15 µC, find the distance from q1 where q2 experiences a net electrostatic force of zero. (The charges q1 and q3 remain separated by a fixed distance of 2d=38 cm.)

As I stare at this, it seems like there should be an easier way than the way I did it:

$$\begin{array}{l} k\frac{{q_{1} \,q_2 }}{{d_{1,2}^2 }} = k\frac{{q_{2} \,q_3 }}{{d_{2,3}^2 }} \\ \\ \rlap{--} k\frac{{q_{1} \,q_2 }}{{d_{1,2}^2 }} = \rlap{--} k\frac{{q_{2} \,q_3 }}{{d_{2,3}^2 }} \\ \\ \frac{{q \times 2q}}{{d_{1,2}^2 }} = \frac{{2q \times 3q}}{{d_{2,3}^2 }} \\ \\ \frac{{2q^2 }}{{d_{1,2}^2 }} = \frac{{6q^2 }}{{d_{2,3}^2 }} \\ \\ \frac{{2\rlap{--} q^{\rlap{--} 2} }}{{d_{1,2}^2 }} = \frac{{6\rlap{--} q^{\rlap{--} 2} }}{{d_{2,3}^2 }} \\ \\ \frac{2}{{d_{1,2}^2 }} = \frac{6}{{d_{2,3}^2 }} \\ \\ d_{2,3} = 0.38 - d_{1,2} \\ \\ \frac{2}{{d_{1,2}^2 }} = \frac{6}{{\left( {0.38 - d_{1,2} } \right)^2 }} \\ \\ \frac{2}{{d_{1,2}^2 }} = \frac{6}{{\left( {0.38 - d_{1,2} } \right)\left( {0.38 - d_{1,2} } \right)}} \\ \\ \frac{2}{{d_{1,2}^2 }} = \frac{6}{{0.38^2 - 0.76d_{1,2} + d_{1,2}^2 }} \\ \\ 2\left( {0.38^2 - 0.76d_{1,2} + d_{1,2}^2 } \right) = 6d_{1,2}^2 \\ \end{array}$$

$$\[ \begin{array}{l} 0.38^2 - 0.76d_{1,2} = 2d_{1,2}^2 \\ \\ - 2d_{1,2}^2 - 0.76d_{1,2} - 0.38^2 = 0 \\ \\ d_{1,2} = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ \\ d_{1,2} = \frac{{ - \left( { - 0.76} \right) \pm \sqrt {\left( { - 0.76} \right)^2 - 4 \times - 2 \times - 0.38^2 } }}{{2 \times - 2}} \\ \\ d_{1,2} = - 0.51908965 \\ d_{1,2} = 0.13908965 \\ \end{array}$$
With 0.139 m or (13.9 cm) being the correct answer

 

[arunbg]

Ahh, the same old story of knowing too much math.
From $$\frac{2}{{d_{1,2}}^2}=\frac{6}{(0.38-{d_{1,2}})^2}$$

you could have simply divided both sides by 2 and taken the square root.
No quadratics involved.

 

[tony873004]

Thanks. I've never been accused of knowing too much math :).

Your way was simpler.

 

[arunbg]

You're welcome ;)