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(C) Let the distance from the charges to point $P$ be $r_1$ and $r_2$. From the geometry,$r_1 = \sqrt{4^2 + 2^2} = \sqrt{20} 	ext{ cm}$ and $r_2 = \s

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$5$

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(C) Let the distance from the charges to point $P$ be $r_1$ and $r_2$. From the geometry,$r_1 = \sqrt{4^2 + 2^2} = \sqrt{20} 	ext{ cm}$ and $r_2 = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 	ext{ cm}$.The electric field due to $q_2$ at $P$ is $E_1 = \frac{K q_2}{r_1^2} = \frac{K q_2}{20}$. The component along the $Y$-axis is $E_{1y} = E_1 \cos \beta = \frac{K q_2}{20} \cdot \frac{4}{\sqrt{20}}$.The electric field due to $q_3$ at $P$ is $E_2 = \frac{K q_3}{r_2^2} = \frac{K q_3}{25}$. The component along the $Y$-axis is $E_{2y} = E_2 \cos 	heta = \frac{K q_3}{25} \cdot \frac{4}{5}$.For the net electric field along the $Y$-axis to be zero,the magnitudes of these components must be equal: $\frac{K q_2}{20} \cdot \frac{4}{\sqrt{20}} = \frac{K q_3}{25} \cdot \frac{4}{5}$.Simplifying,$\frac{q_2}{20 \sqrt{20}} = \frac{q_3}{125} \Rightarrow \frac{q_2}{q_3} = \frac{20 \sqrt{20}}{125} = \frac{4 \sqrt{20}}{25} = \frac{4 \cdot 2 \sqrt{5}}{25} = \frac{8 \sqrt{5}}{25} = \frac{8}{5 \sqrt{5}}$.Comparing this with $\frac{8}{5 \sqrt{x}}$,we get $x = 5$.

If the net electric field at point $P$ along the $Y$-axis is zero,then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$,where $x = . . . . . .$

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