Two point charges -10 mu C and +5 mu C are si | vedclass

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(C) Let the charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$. Since the charges have opposite signs,the electric

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$x=2(\sqrt{2}+1) \ m$

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(C) Let the charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$. Since the charges have opposite signs,the electric field can only be zero at a point outside the line segment joining the charges,specifically on the side of the smaller magnitude charge $(q_2)$. Let the neutral point be at distance $d$ from $q_2$ in the positive $x$-direction. The electric field due to $q_1$ and $q_2$ at this point must be equal in magnitude: $\frac{k|q_1|}{(d + \sqrt{2})^2} = \frac{k|q_2|}{d^2}$ $\frac{10}{(d + \sqrt{2})^2} = \frac{5}{d^2}$ $2d^2 = (d + \sqrt{2})^2$ Taking the square root on both sides: $\sqrt{2}d = d + \sqrt{2}$ $d(\sqrt{2} - 1) = \sqrt{2}$ $d = \frac{\sqrt{2}}{\sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} + 1)}{2 - 1} = 2 + \sqrt{2} \ m$. The coordinate $x$ of this point is $x_2 + d = \sqrt{2} + (2 + \sqrt{2}) = 2 + 2\sqrt{2} = 2(\sqrt{2} + 1) \ m$.

Two point charges $-10 \mu C$ and $+5 \mu C$ are situated on the $X$-axis at $x=0$ and $x=\sqrt{2} \ m$. The point along the $X$-axis where the electric field becomes zero is:

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