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

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$\frac{- Q}{3 \sqrt{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

$\frac{- 2 Q}{3 \sqrt{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

$\frac{2 Q}{3 \sqrt{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

$\frac{Q}{3 \sqrt{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

answer is B.

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Detailed Solution

We can replace – Q charge at origin by +Q and -2Q. Now due to +Q charge at every corner of cube. Electric field at centre of sube is zero so now net electric field at centre is only to -2Q charge at Origin.

$\bar{E} = \frac{K q \bar{r}}{r^{3}} = \frac{1 (- 2 Q) \frac{a}{2} (\hat{x} + \hat{y} + \hat{z})}{4 π ε_{0} {(\frac{a}{2} \sqrt{3})}^{3}} \bar{E} = \frac{- 2 Q (\hat{x} + \hat{y} + \hat{z})}{3 \sqrt{3} π a^{2} ε_{0}}$