Question

Q) A straight, nonconducting plastic wire 8.50 cm long carries a charge density of +175 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 cm directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 cm directly above its center.

Short answer

Answer:

(a) The magnitude of the electric field at point a point 6.00cm directly above its midpoint is$\text{3.04}\times {\text{10}}^4 \text{N/C}$and the direction is in positive x direction.

(b) The magnitude of the electric field at point a point 6.00cm directly above its center is $\text{3.46}\times {{10}^{-}}^4\text{N/C}$and the direction of electric field is in +x direction .

Step-by-step solution

Formula related to the electric field of the ring

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=$\frac{1}{4\pi {\epsilon }_{\circ }}=9.0\times {10}^9 {\text{N.m}}^2/{\text{C}}^2$, Q=charge on the ring and a= radius of the ring

Calculating the radius of the ring and charge on the ring

The length a is half the total length

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

The total charge is

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

Substitute the values to find E:

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

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

If the wire is forming a ring

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=$\frac{1}{4\pi {\epsilon }_{\circ }}=9.0\times {10}^9 {\text{N.m}}^2/{\text{C}}^2$, Q=charge on the ring and a= radius of the ring

The radius a is given by

$$\begin{gathered} 2\pi a=L \\ a=\frac{8.5\times {10}^{-2}}{2\pi } \\ =1.35\times {10}^{-2} \text{m} \end{gathered}$$

Substitute the values to find E:

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

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