91Ó°ÊÓ

The uniform line charge is present in the form of a square ring. The electric force exerted by the ring at a distance z on the z-axis, just above the center of the square loop, has to be evaluated.

Electric force exerted by charge on charge is proportional to the product of the two charge and inversely proportional to the square of the distance between them as,

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\frac{\mathbf{qQ}}{{\mathbf{r}}^{\mathbf{2}}}$.

The uniform line charge, present in the form of a square ring.is shown in following figure, along with the components of electric field at the point on the z axis.

The expression of the electric field, due to the straight segment of the square loop, at a distance $r$on the z axis, lying at its center can be written as,

$E^{\text{'}}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{2\mathrm{λ}L}{r\sqrt{r+L^2}}\right)$ ……. (1)

Here $\mathrm{λ}$is the line charge density, $L$is the length of the side, ${\mathrm{Î\mu }}_0$is the permittivity of free space.

It is given that the side of square loop is $a$, such that $2L=a$, then we can write

$L=\frac{a}{2}$

From the above figure, from the right triangle, using Pythagoras theorem we can write

$\begin{array}{l}{r}^2=z^2+\frac{a^2}{4}\\ r=\sqrt{z^2+\frac{a^2}{4}}\end{array}$

Theelectric field at the point, lying on the center of the square loop at a distanceon the z axis, is obtained as follows:

Substitute $\frac{a}{2}$for $L$, and $\sqrt{z^2+\frac{a^2}{4}}$for $r$into equation (1)

$$\begin{gathered} E^{\text{'}}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{2\mathrm{λ}\left({\displaystyle \frac{a}{2}}\right)}{\left(\sqrt{z^2+{\displaystyle \frac{a^2}{4}}}\right)\sqrt{{\left(z^2+\frac{a^2}{4}\right)}^2+{\left({\displaystyle \frac{a}{2}}\right)}^2}}\right) \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{a\mathrm{λ}}{\left(\sqrt{z^2+{\displaystyle \frac{a^2}{4}}}\right)\sqrt{z^2+\frac{a^2}{4}+\frac{a^2}{4}}}\right) \\ =\frac{a\mathrm{λ}}{\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}} \end{gathered}$$

The resultant electric field is the sum of electric field due to all the sides of the square loop .from the diagram it can be inferred that the horizontal components are equal and opposite to each other .Thus the horizontal component cancels out. So, the resultant electric field vector is in vertical direction, as

$E=4E^{\text{'}}\cos \mathrm{θ}$ …… (2)

From the diagram, we can write

$\cos \mathrm{θ}=\frac{z}{r}$

Substitute $\sqrt{z^2+\frac{a^2}{4}}\mathrm{for}r$into the equation.

$\cos \mathrm{θ}=\frac{z}{\sqrt{z^2+\frac{a^2}{4}}}$

Substitute $\frac{z}{\sqrt{z^2+\frac{a^2}{4}}}\mathrm{for}cos\mathrm{θ},\frac{a\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}}\mathrm{for}{E}^{\text{'}}$into equation (2)

$$\begin{gathered} E=4\left(\frac{a\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}}\right)\left(\frac{z}{\sqrt{z^2+\frac{a^2}{2}}}\right) \\ =\frac{4a\mathrm{λ}z}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(z^2+\frac{a^2}{2}\right)\sqrt{z^2+\frac{a^2}{4}}} \end{gathered}$$

Therefore, the resultant electric field at a distance z above the center line of the square loop is $E=\frac{4a\mathrm{λ}z}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(z^2+\frac{a^2}{2}\right)\sqrt{z^2+\frac{a^2}{4}}}\hat{z}$.