Two charges 9e and 3e are placed at a distanc | vedclass

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Question

(A) Let the electric field be zero at a distance $x_1$ from the charge $Q_1 = 9e$ and at a distance $x_2$ from the charge $Q_2 = 3e$, where $x_1 + x_2

Vedclass · Accepted Answer

$\frac{r}{1 + \sqrt{1/3}}$

Vedclass · Answer

(A) Let the electric field be zero at a distance $x_1$ from the charge $Q_1 = 9e$ and at a distance $x_2$ from the charge $Q_2 = 3e$, where $x_1 + x_2 = r$.At the null point, the magnitudes of the electric fields produced by both charges must be equal:$\frac{k Q_1}{x_1^2} = \frac{k Q_2}{x_2^2}$$\frac{9e}{x_1^2} = \frac{3e}{x_2^2}$$\frac{x_1^2}{x_2^2} = \frac{9e}{3e} = 3$$\frac{x_1}{x_2} = \sqrt{3} \implies x_1 = x_2 \sqrt{3}$Since $x_1 + x_2 = r$, we have $x_2 \sqrt{3} + x_2 = r \implies x_2(1 + \sqrt{3}) = r \implies x_2 = \frac{r}{1 + \sqrt{3}}$.Now, $x_1 = r - x_2 = r - \frac{r}{1 + \sqrt{3}} = r \left( \frac{1 + \sqrt{3} - 1}{1 + \sqrt{3}} \right) = \frac{r \sqrt{3}}{1 + \sqrt{3}}$.Dividing the numerator and denominator by $\sqrt{3}$, we get $x_1 = \frac{r}{1 + \frac{1}{\sqrt{3}}} = \frac{r}{1 + \sqrt{1/3}}$.

Two charges $9e$ and $3e$ are placed at a distance $r$ from each other. At what distance from the $9e$ charge will the electric field be zero?

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