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.)(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]

\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}

[/tex]

[tex]

\[

\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}

[/tex]

With 0.139 m or (13.9 cm) being the correct answer

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