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(D) Let the square lie in the $xy$-plane with its center at the origin. The vertical axis is the $z$-axis. The four corners are at $(\pm a/\sqrt{2}, \

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$-4q$

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(D) Let the square lie in the $xy$-plane with its center at the origin. The vertical axis is the $z$-axis. The four corners are at $(\pm a/\sqrt{2}, \pm a/\sqrt{2}, 0)$.Three corners have charge $q$,and one corner has charge $Q+q$.The electric field component along the $z$-axis due to a charge $q_i$ at distance $r_i$ from the axis is $E_{z,i} = \frac{k q_i z}{(r_i^2 + z^2)^{3/2}}$.For the three charges $q$,the sum of the vertical components is $E_{z,1+2+3} = \frac{3kqz}{(r^2 + z^2)^{3/2}}$.For the fourth charge $(Q+q)$,the vertical component is $E_{z,4} = \frac{k(Q+q)z}{(r^2 + z^2)^{3/2}}$.For the net field along the $z$-axis to be zero,$E_{z,1+2+3} + E_{z,4} = 0$.$3kq + (Q+q) = 0 \implies Q+q = -3q \implies Q = -4q$.

For the given arrangement,where four charges are fixed at the corners of a square as shown,find the value of the additional charge $Q$ to be placed on one of the vertices so that the component of the net electric field along the vertical symmetric axis is zero at every point on the vertical axis.

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