Question

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

Answer 1

The correct option is B Only $x=-\sqrt{2}a$

By intuition we know that it will be at two point i.e., A and B.

Let distance of A from $(a,0)$ be x and distance of B from $(2a,0)$ be y.

Point A:

$\frac{k(+Q)}{x^2}+\frac{k(-2Q)}{(a+x{)}^2}=0$

$\frac{1}{x^2}-\frac{2}{(a+x{)}^2}=0$

$(a+x{)}^2=2x^2$

$a^2+x^2+2ax=2x^2$

$x^2-2ax-a^2=0$

$x=\frac{2a\pm \sqrt{4a^2+4a^2}}{2}=\frac{2a\pm \sqrt{8a^2}}{2}=\frac{2a\pm 2\sqrt{2a}}{2}$

$x=(1\pm \sqrt{2})a$

$\Rightarrow x=(1+\sqrt{2})a$

Point B:

$\frac{K(+Q)}{(a+y{)}^2}+\frac{K(-2Q)}{y^2}=0$

$\frac{1}{(a+y{)}^2}=\frac{2}{y^2}$

$y^2=2(a+y{)}^2$

$y^2=2a^2+2y^2+4ay$

$y^2+4ay+2a^2=0$

$y=\frac{-4a\pm \sqrt{16a^2-4a^2}}{2}=\frac{-4a\pm \sqrt{8a^2}}{2}=\frac{-4a\pm 2\sqrt{2a}}{2}$

$y=(-2\pm \sqrt{2})a$

$\Rightarrow y=$ Not feasible as both are negative.

Thus, as x is found, accordingly the point on x-axis will be $(-\sqrt{2a},0)$.