Two electric charges 9e and 3e are placed at | vedclass

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(B) Let the point where the electric field is zero be at a distance $x$ from the charge $9e$. At this point,the magnitude of the electric field due to

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$\frac{r}{(1 + \sqrt{1/3})}$ from $9e$ charge

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(B) Let the point where the electric field is zero be at a distance $x$ from the charge $9e$. At this point,the magnitude of the electric field due to both charges must be equal: $\frac{k(9e)}{x^2} = \frac{k(3e)}{(r-x)^2}$ Taking the square root on both sides: $\frac{3}{x} = \frac{\sqrt{3}}{r-x}$ $3(r-x) = x\sqrt{3}$ $3r - 3x = x\sqrt{3}$ $3r = x(3 + \sqrt{3})$ $x = \frac{3r}{3 + \sqrt{3}}$ Dividing the numerator and denominator by $3$: $x = \frac{r}{1 + \frac{\sqrt{3}}{3}} = \frac{r}{1 + \frac{1}{\sqrt{3}}} = \frac{r}{(1 + \sqrt{1/3})}$ Thus,the distance from the $9e$ charge is $\frac{r}{(1 + \sqrt{1/3})}$.

Two electric charges $9e$ and $3e$ are placed at a distance $r$ from each other. The point where the electric field intensity is zero is at a distance of .......

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