Two point charges -q and +q/2 are situated at | vedclass

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(C) Let the electric field vanish at a point $P$ on the $X$-axis at a distance $x$ from the origin. The electric field due to charge $-q$ at the origi

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$x = \frac{\sqrt{2}a}{\sqrt{2} - 1}$

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(C) Let the electric field vanish at a point $P$ on the $X$-axis at a distance $x$ from the origin. The electric field due to charge $-q$ at the origin is $E_1 = \frac{kq}{x^2}$ (directed towards the origin).The electric field due to charge $+q/2$ at $(a, 0, 0)$ is $E_2 = \frac{k(q/2)}{(x - a)^2}$ (directed away from the origin).For the net electric field to be zero,the magnitudes must be equal: $E_1 = E_2$.$\frac{kq}{x^2} = \frac{kq/2}{(x - a)^2}$$2(x - a)^2 = x^2$Taking the square root on both sides: $\sqrt{2}(x - a) = x$ (considering the region $x > a$ where the fields can cancel).$\sqrt{2}x - \sqrt{2}a = x$$(\sqrt{2} - 1)x = \sqrt{2}a$$x = \frac{\sqrt{2}a}{\sqrt{2} - 1}$

Two point charges $-q$ and $+q/2$ are situated at the origin $(0, 0, 0)$ and at the point $(a, 0, 0)$ respectively. The point along the $X$-axis where the electric field vanishes is:

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