Question

A wire having uniform linear charge density € is
placed in circular form in 3 quadrants with different
polarities as shown in the figure. Let Eo be the
electric field at O. Then the angle made by Eo
with the line along y-axis is​

Answer 1

Answer:

$(3) tan^{-1}(\frac{1}{3})$

Explanation:

1)  Since, linear charge density of wire is uniform.

Let the electric field by one quarter of Wire be E.

In one quarter wire(say negative polarity wire), all charges are symmetric about angle bisector of that quadrant.

=> Electric field in one quadrant is along angle bisector(45 degree) of that quadrant.

Then, Similarly on other negative polarity , Electric field is along that bisector (45 degree)

2) Now, for the case of electric field due to positive polar wire , is along the direction of  electric field of right negative polar wire.

That is, field becomes 2E

Our net figure is attached.

Let the $\vec{E_o}$ makes angle $\theta$ with 2E and $\alpha$ with y-axis.

$=>tan(\theta)=\frac{E}{2E}=\frac{1}{2}$

And,

$tan(\alpha+\theta)=\frac{\pi}{4}\\ \\=>\frac{tan(\alpha)+tan(\theta)}{1-tan(\theta)tan(\alpha)}=1\\ \\=>\frac{tan(\alpha)+1/2}{1-1/2tan(\alpha)}=1\\ \\=>tan(\alpha)+1/2=1-1/2tan(\alpha)\\ \\=>tan(\alpha)=\frac{1}{3} \\ \\=>\alpha=tan^{-1}(1/3)$

Hence, Required angle is $tan^{-1}(1/3)$.