Question

A cube of side $l$ has point charges $(+q)$ at each of its vertices except at the origin where the charge is $(-q)$ as shown in the figure. The electric field at the centre of the cube is

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

Answer 1

The correct option is D $\frac{-2q}{3\sqrt{3}\pi {ϵ}_0{l}^2}(\hat{x}+\hat{y}+\hat{z})$


Let us assume there is charge $-2q$ and $+q$ at the vertex $4$. So, total charge at this vertex remain the same, which is $-q$.

Therefore, the net electric field due to all the eight $+q$ charges at the centre of the cube.

But there will be electric field at the centre of cube due to the charge $-2q$.

Hence,

$\overrightarrow{E}=\frac{-2q\overrightarrow{r}}{4\pi {ϵ}_o{r}^3}$

Here, $\overrightarrow{r}=\frac{l}{2}\hat{x}+\frac{l}{2}\hat{y}+\frac{l}{2}\hat{z}$

So, $r=\sqrt{(l/2{)}^2+(l/2{)}^2+(l/2{)}^2}=\frac{\sqrt{3}l}{2}$

$\therefore \overrightarrow{E}=\frac{-2q}{4\pi {ϵ}_o{\left(\sqrt{3}l/2\right)}^3}(\frac{l}{2}\hat{x}+\frac{l}{2}\hat{y}+\frac{l}{2}\hat{z})$

$\Rightarrow \overrightarrow{E}=\frac{-2q}{3\sqrt{3}\pi {ϵ}_o{l}^2}(\hat{x}+\hat{y}+\hat{z})$