Find the value of q so that the net electrost | vedclass

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(C) Let the side length of the square be $a$. Consider the $10 \mu C$ charge at the top-right corner. The forces acting on it are:$1$. Force due to th

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$-\frac{5}{\sqrt{2}} \mu C$

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(C) Let the side length of the square be $a$. Consider the $10 \mu C$ charge at the top-right corner. The forces acting on it are:$1$. Force due to the $10 \mu C$ charge at the bottom-left corner: $F_1 = \frac{k (10 \mu C)^2}{(\sqrt{2} a)^2}$ directed along the diagonal.$2$. Forces due to the two $q$ charges at the other vertices: $F_2 = \frac{k q (10 \mu C)}{a^2}$ directed along the sides.For the net force to be zero,the resultant of the two $F_2$ forces must be equal and opposite to $F_1$.The resultant of the two $F_2$ forces is $\sqrt{2} F_2 = \sqrt{2} \frac{k q (10 \mu C)}{a^2}$.Equating the magnitudes: $\sqrt{2} \frac{k q (10 \mu C)}{a^2} = - \frac{k (10 \mu C)^2}{2 a^2}$.Solving for $q$: $q = - \frac{10 \mu C}{2 \sqrt{2}} = - \frac{5}{\sqrt{2}} \mu C$.

Find the value of $q$ so that the net electrostatic force on the $10 \mu C$ charge placed at a vertex of the square is zero. The charges are arranged as shown in the figure.

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