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 center 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)

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

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 center if the cube is zero. Now net electric field at the center is only due to $-2Q$ charge at the origin.

$\overrightarrow{E}=\frac{kq\overrightarrow{r}}{r^3}=\frac{1(-2Q)\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})}{4\pi {\epsilon }_0{\left(\frac{a}{2}\sqrt{3}\right)}^3}$
$\overrightarrow{E}=\frac{-2Q(\hat{x}+\hat{y}+\hat{z})}{3\sqrt{3}\pi a^2{\epsilon }_0}$