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(D) Let the four charges $q$ be placed at $(\pm L/2, \pm L/2, 0)$. The distance of any charge from a point $(0, 0, z)$ on the $Z$-axis is $r = \sqrt{(

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(D) Let the four charges $q$ be placed at $(\pm L/2, \pm L/2, 0)$. The distance of any charge from a point $(0, 0, z)$ on the $Z$-axis is $r = \sqrt{(L/2)^2 + (L/2)^2 + z^2} = \sqrt{L^2/2 + z^2}$.The electric field due to one charge at distance $r$ is $E_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$.The component of this field along the $Z$-axis is $E_z = E_1 \cos	heta$,where $\cos	heta = z/r$.Since there are four charges,the total electric field is $E = 4 	imes E_z = 4 	imes \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} 	imes \frac{z}{r} = \frac{q z}{\pi\epsilon_0 (L^2/2 + z^2)^{3/2}}$.To find the maximum,set $dE/dz = 0$:$\frac{d}{dz} [z (L^2/2 + z^2)^{-3/2}] = (L^2/2 + z^2)^{-3/2} + z(-3/2)(L^2/2 + z^2)^{-5/2}(2z) = 0$.$(L^2/2 + z^2) - 3z^2 = 0 \implies L^2/2 = 2z^2 \implies z^2 = L^2/4 \implies z = L/2$.

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