A point charge +q is placed at the origin. A | vedclass

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(A) Let the point $P$ be at a distance $x$ from the origin $(0, 0, 0)$ along the $x$-axis,where the net electric field is zero.At point $P$,the electr

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$(d/4, 0, 0)$

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(A) Let the point $P$ be at a distance $x$ from the origin $(0, 0, 0)$ along the $x$-axis,where the net electric field is zero.At point $P$,the electric field due to charge $+q$ at the origin must be equal in magnitude and opposite in direction to the electric field due to charge $+9q$ at $(d, 0, 0)$.Thus,$\frac{kq}{x^2} = \frac{k(9q)}{(d-x)^2}$.Taking the square root on both sides,we get $\frac{1}{x} = \frac{3}{d-x}$.Solving for $x$: $d - x = 3x$,which gives $4x = d$,or $x = d/4$.Therefore,the coordinate of the point $P$ is $(d/4, 0, 0)$.

$A$ point charge $+q$ is placed at the origin. $A$ second point charge $+9q$ is placed at $(d, 0, 0)$ in a Cartesian coordinate system. The point in between them where the electric field vanishes is:

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