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 cube is:

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

Answer: (B)

We can replace $-Q$ charge at origin by $+Q$ and $-2Q.$
Now due to $+Q$ charge at every corner of cube. Electric field at center of cube is zero so now net electric field at center is only due to $-2Q$ charge at origin.
$\vec{E}=\frac{kq\vec{r}}{r^3}=\frac{1(-2Q)\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})}{4\pi {\epsilon }_0{\left(\frac{a}{2}\sqrt{3}\right)}^3}$
$\vec{E}=\frac{-2Q(\hat{x}+\hat{y}+\hat{z})}{3\sqrt{3}\pi a^2{\epsilon }_0}$