Two point charges ( + Q) and ( - 2Q) are fixe | vedclass

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Question

(b) Suppose electric field is zero at a point $P$ lies at a distance d from the charge $+ Q$.
At $P$ $\frac{{kQ}}{{{d^2}}} = \frac{{k(2Q)}}{{{{(a + d

Vedclass · Accepted Answer

Only $x = - \sqrt 2 a$

Vedclass · Answer

(b) Suppose electric field is zero at a point $P$ lies at a distance d from the charge $+ Q$.
At $P$ $\frac{{kQ}}{{{d^2}}} = \frac{{k(2Q)}}{{{{(a + d)}^2}}}$
$==>$ $\frac{1}{{{d^2}}} = \frac{2}{{{{(a + d)}^2}}}$ $==>$ $d = \frac{a}{{(\sqrt 2 - 1)}}$
Since $d > a$ i.e. point $P$ must lies on negative $x$-axis as shown at a distance $x$ from origin hence $x = d - a$ $ = \frac{a}{{(\sqrt 2 - 1)}} - a = \sqrt 2 \,a.$ Actually $P$ lies on negative $x$-axis so $x = - \sqrt 2 \,a$

Two point charges $( + Q)$ and $( - 2Q)$ are fixed on the $X-$axis at positions $a$ and $2a$ from origin respectively. At what positions on the axis, the resultant electric field is zero

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