Question #a20f5

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Question #a20f5

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Answer 1 · 2016-07-29T13:53:56

$\frac{1}{3}$ $1/3$ meters to the right of $q 1$ $q1$ and $3$ $3$ meters to the left of $q_{1}$ $q_1$ (assuming $q_{2}$ $q_2$ is to the right of $q_{1}$ $q_1$ )

Explanation:

Electric field from $q_{1}$ $q_1$ to cancel $q_{2}$ $q_2$ would mean:
$E_{1} = E_{2}$ $E_1=E_2$

$k \frac{q_{1}}{r^{2}} = k \frac{q_{2}}{{(1 - r)}^{2}}$ $kq_1/r^2=kq_2/(1-r)^2$
$r$ $r$ is the distance from $q_{1}$ $q_1$ .

Rearrange the equation and you have a quadratic equation:
$3 r^{2} + 2 r - 1 = 0$ $3r^2+2r-1=0$

Find the roots and you have $\frac{1}{3} m$ $1/3m$ (to the right) and $- 3 m$ $-3m$ (to the left).

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Question #a20f5

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