Question

What will be the magnitude of electric field at point $O$ as shown in figure? Each side of the figure is $l$ and perpendicular to each other ?

• A. $\frac{1}{4\pi {ϵ}_0}\frac{q}{l^2}$
• B. $\frac{1}{4\pi {ϵ}_0}\frac{q}{\left(2l^2\right)}(2\sqrt{2}-1)$
• C. $\frac{q}{4\pi {ϵ}_0{\left(2l\right)}^2}$
• D. $\frac{1}{4\pi {ϵ}_0}\frac{2q}{2l^2}\left(\sqrt{2}\right)$

Answer 1

The correct option is B $\frac{1}{4\pi {ϵ}_0}\frac{q}{\left(2l^2\right)}(2\sqrt{2}-1)$

Given,
If we talk about the charges at points $B$ and $G$ , the net electric field due to them, at point $O$ will be towards $B$, as $2q>q.$
Similarly, if we talk about the charges at points $C$ and $F$ , the net electric field due to them, at point $O$ will be towards $F$, as $2q>q.$
Again, if we talk about rest of the charges at different positions,
there will be no electric field at $O$ in the alignment of $AH$ as the magnitude of electric field due to charges at $A$ and $H$ are equal, so, they will cancle out each other.
There will be net electric field at point $O$ along $OD$, as $2q>q$
Let us represent this situation pictorially,

Where, $E_1,E_2\text{and}{E}_3$ are the net electric field at $O$ as per the mentioned above conditions.
Thus,
$E_1=\frac{2kq}{l^2}-\frac{kq}{l^2}=\frac{kq}{l^2}$
$E_2=\frac{2kq}{l^2}-\frac{kq}{l^2}=\frac{kq}{l^2}$
$E_3=\frac{2kq}{{\left(\sqrt{2}l\right)}^2}-\frac{kq}{{\left(\sqrt{2}l\right)}^2}=\frac{kq}{2l^2}$
After resolving these electric fields and solving it, we get the magnitude of resultant electric field at $O$ as
$E_R=\sqrt{2}\times \frac{kq}{l^2}(1-\frac{1}{2\sqrt{2}})=\frac{kq}{2l^2}(2\sqrt{2}-1)$
Or, $E_R=\frac{1}{4\pi {ϵ}_0}\frac{q}{\left(2l^2\right)}(2\sqrt{2}-1)$
Hence. option (b) is correct.