A cube of side $a$ has point charges $+Q$ located at each of its vertices except at the origin where the charge is $-Q$. The electric field at the centre of cube is:
Answer: (C) $\frac{-2Q}{3\sqrt3\pi\varepsilon_0a^2}(\hat x+\hat y+\hat z)$
Write the charge at the origin as $+Q$ plus an extra $-2Q$. Eight equal charges $+Q$ at the corners give zero field at the centre by symmetry, so only the $-2Q$ at the origin matters.
The centre is at $\frac{a}{2}(1,1,1)$, at distance $r = \frac{\sqrt3 a}{2}$, so $r^2 = \frac{3a^2}{4}$. The field of a negative charge points towards it, i.e. along $-\frac{(\hat x+\hat y+\hat z)}{\sqrt3}$:
$$\vec E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2Q}{3a^2/4}\cdot\left(-\frac{\hat x+\hat y+\hat z}{\sqrt3}\right) = \frac{-2Q}{3\sqrt3\pi\varepsilon_0a^2}(\hat x+\hat y+\hat z)$$
Solution by Sreeraj P, M.Sc Physics