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

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(C) Let the two charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$. For the electric field to be zero,the point mu

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

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(C) Let the two charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$. For the electric field to be zero,the point must be closer to the charge with the smaller magnitude. Since $|q_2| Let the distance of point $P$ from $q_2$ be $x$. The position of $P$ is $x_P = \sqrt{2} + x$. The electric field due to $q_1$ at $P$ is $E_1 = \frac{k|q_1|}{(\sqrt{2}+x)^2}$ (directed towards the origin). The electric field due to $q_2$ at $P$ is $E_2 = \frac{k|q_2|}{x^2}$ (directed away from the origin). Setting $E_1 = E_2$: $\frac{10}{x^2 + 2\sqrt{2}x + 2} = \frac{5}{x^2}$ $2x^2 = x^2 + 2\sqrt{2}x + 2$ $x^2 - 2\sqrt{2}x - 2 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{2\sqrt{2} \pm \sqrt{8 - 4(1)(-2)}}{2} = \frac{2\sqrt{2} \pm \sqrt{16}}{2} = \sqrt{2} \pm 2$. Since $x$ must be positive,$x = \sqrt{2} + 2$. The coordinate of point $P$ is $x_P = \sqrt{2} + x = \sqrt{2} + (\sqrt{2} + 2) = 2\sqrt{2} + 2 = 2(\sqrt{2} + 1) \ m$.

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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