91Ó°ÊÓ

Electric Field of a Line Segment at distance x above the midpoint of a straight line segment:

$\mathbf{E}\mathbf{=}\frac{\mathbf{kQ}}{\mathbf{x}{\sqrt{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}}}^{}}$

where k=$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{∘}}=9.0×{10}^9 {\text{N.m}}^2/{\text{C}}^2$, Q=charge on the ring and a= radius of the ring

The length a is half the total length

$$\begin{gathered} a=\frac{L}{2} \\ =\frac{8.5}{2}×{10}^{-2} \\ =4.25×{10}^{-2}\text{m/s} \end{gathered}$$

The total charge is

$$\begin{gathered} Q=\mathrm{λ}L \\ =175×{10}^{-9}×8.5×{10}^{-2} \\ =1.49×{10}^{-8}\text{C} \end{gathered}$$

Substitute the values to find E:

$$\begin{gathered} E=\frac{9×{10}^9×\text{1.49}×{{10}^{-}}^8}{0.06\sqrt{{(0.06)}^2+{(0.425)}^2}} \\ =\text{3.04}×{\text{10}}^4 \text{N/C} \end{gathered}$$

Since the charge is positive so the field line is in positive x direction

If the wire is formed to a loop then the electric field at a distance x on the axis of a uniformly charged ring is:

$\mathbf{E}\mathbf{=}\frac{\mathbf{kQx}}{{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$

where k=$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{∘}}=9.0×{10}^9 {\text{N.m}}^2/{\text{C}}^2$, Q=charge on the ring and a= radius of the ring

The radius a is given by

$$\begin{gathered} 2\mathrm{π}a=L \\ a=\frac{8.5×{10}^{-2}}{2\mathrm{π}} \\ =1.35×{10}^{-2} \text{m} \end{gathered}$$

Substitute the values to find E:

$$\begin{gathered} E=\frac{9×{10}^9×\text{1.49}×{{10}^{-}}^9×\text{6}×{{10}^{-}}^2}{{({0.06}^2+{0.0135}^2)}^{3/2}} \\ =\text{3.46}×{{10}^{-}}^4\text{N/C} \end{gathered}$$

Since the charge is positive so the field line is in +x direction.