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 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

If we consider two point charges +Q and –2Q at origin, then +Q charge along with other seven +Q charges will produce zero net electric field at centre of cube due to symmetry. Now only remaining charge is –2Q at center.

So net electric field at centre will only be due to -2Q charge.

$\begin{array}{l} \vec{E} = \frac{k q}{r^{3}} \vec{r} here, \vec{r} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k} \\ \Rightarrow \vec{E} = \frac{k (- 2 Q)}{{(\frac{a \sqrt{3}}{2})}^{3}} (\frac{a}{2} \hat{i} + \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}) \\ \Rightarrow \vec{E} = \frac{- 2 Q}{3 \sqrt{3} π ε_{0} a^{2}} (\hat{i} + \hat{j} + \hat{k}) \\ We can replace unit vectors \hat{i}, \hat{j}, \hat{k} with \hat{x}, \hat{y}, \hat{z} \end{array}$ .