Q. 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
A) Only $x = \sqrt{2}a$
B) Only $x = - \sqrt{2}a$
C) Both $x = \pm \sqrt{2}a$
D) $x = \frac{3a}{2}$ only
Explanation
(b) Suppose electric field is zero at a point $P$ lies at a distance $d$ from the charge $+ Q$
At $P\ \frac{kQ}{d^{2}} = \frac{k(2Q)}{(a + d)^{2}}$
$\Rightarrow \frac{1}{d^{2}} = \frac{2}{(a + d)^{2}} \Rightarrow d = \frac{a}{(\sqrt{2} - 1)}$
Since $d > a\ i.e.$ point $P$ must lies on negative $x$-axis as shown at a distance $x$ from origin hence $x = d - a = \frac{a}{\left( \sqrt{2} - 1 \right)} - a = \sqrt{2}a.$ Actually $P$ lies on negative $x$-axis so $x = - \sqrt{2}a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.