Two charges q_1 = +6q and q_2 = -3q are place | vedclass

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(A) Let the proton (charge $+e$) be placed at a distance $x$ from $q_1$ on the $x$-axis. Since $q_2$ is at distance $L$ from $q_1$,the distance of the

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$\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right) L$

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(A) Let the proton (charge $+e$) be placed at a distance $x$ from $q_1$ on the $x$-axis. Since $q_2$ is at distance $L$ from $q_1$,the distance of the proton from $q_2$ is $(x - L)$.For the proton to be in equilibrium,the net electrostatic force on it must be zero.The force due to $q_1$ is $F_1 = \frac{k q_1 e}{x^2}$ (repulsive,towards the right).The force due to $q_2$ is $F_2 = \frac{k |q_2| e}{(x - L)^2}$ (attractive,towards the left).Equating the magnitudes: $\frac{k (6q) e}{x^2} = \frac{k (3q) e}{(x - L)^2}$.Simplifying: $\frac{6}{x^2} = \frac{3}{(x - L)^2} \implies \frac{2}{x^2} = \frac{1}{(x - L)^2}$.Taking the square root on both sides: $\frac{\sqrt{2}}{x} = \frac{1}{x - L}$.$\sqrt{2}(x - L) = x \implies \sqrt{2}x - \sqrt{2}L = x$.$x(\sqrt{2} - 1) = \sqrt{2}L$.$x = \left(\frac{\sqrt{2}}{\sqrt{2} - 1}\right) L$.

Two charges $q_1 = +6q$ and $q_2 = -3q$ are placed as shown in the figure. $A$ proton is placed on the $x$-axis away from $q_2$. To keep the proton in equilibrium,the distance between $q_1$ and the proton is:

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