Question

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 the cube is :

• A. $\frac{2Q}{3\sqrt{3}\pi {\epsilon }_0{a}^2}(\hat{x}+\hat{y}+\hat{z})$
• B. $\frac{Q}{3\sqrt{3}\pi {\epsilon }_0{a}^2}(\hat{x}+\hat{y}+\hat{z})$
• C. $\frac{-2Q}{3\sqrt{3}\pi {\epsilon }_0{a}^2}(\hat{x}+\hat{y}+\hat{z})$
• D. $\frac{-Q}{3\sqrt{3}\pi {\epsilon }_0{a}^2}(\hat{x}+\hat{y}+\hat{z})$

Answer 1

The correct option is C $\frac{-2Q}{3\sqrt{3}\pi {\epsilon }_0{a}^2}(\hat{x}+\hat{y}+\hat{z})$

We can replace $–Q$ charge at origin by $+Q$ and $–2Q$.

Now, due to $+Q$ charge at every corner of the cube, the electric field at the centre of the cube, is zero.

So, net electric field at centre is only due to $–2Q$ charge at origin.

Position vector of centre of the cube,
$\overrightarrow{r}=\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})$

Also, $r=\frac{\sqrt{3}a}{2}$

Further,
$\overrightarrow{E}=\frac{q\overrightarrow{r}}{4\pi {\epsilon }_0{r}^3}$

$\overrightarrow{E}=\frac{-2Q\times \frac{a}{2}(\hat{x}+\hat{y}+\hat{z})}{4\pi {\epsilon }_0{\left(\frac{\sqrt{3}a}{2}\right)}^3}$

$\overrightarrow{E}=\frac{-2Q}{3\sqrt{3}\pi {\epsilon }_0{a}^2}(\hat{x}+\hat{y}+\hat{z})$