Q5P - Introduction to Electrodynamics

Question

Find the electric field a distance z above the center of a circular loop of radius $\mathit{r}$ (Fig. 2.9) that carries a uniform line charge $\mathit{\lambda }$

                      

Step-by-Step Solution

Answer

The electric field at a distance z above the center of a circular loop 

is

$E=\frac{\lambda }{2{\epsilon }_0}\left(\frac{zr}{{\left(r^2+z^2\right)}^{3/2}\ne }\right)$

Step 1: Describe the given information

The radius of the circuit is $r$.

The uniform line charge is $\lambda$.

Step 2: Define the coulomb’s law

Electric field due to charge $\mathit{q}$ at a distance $\mathit{r}$ is proportional to the charge $\mathit{q}$and inversely proportional to the square of the distance $\mathit{r}$as,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$ 

Step 3: Obtain the electric field above the center of circular loop

The center line of the circular loop of radius $r$ is drawn. At a point z on the center line the edges of the loop make the angle$\theta$, as shown below:

                       

From the above, using Pythagoras theorem in left right triangle, we can write$R^2=r^2+;$

$\cos \theta =\frac{z}{\sqrt{r^2+z^2}}$

It can be considered that the circular loop is made up of symmetrical elements of small lengths $dx$, which are diagonally opposite then the differential field at the point P due to a pair of diagonally opposite elements can be written as

$dE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{R^2}\cos \theta$.

Substitute $r^2+z^2$ for$R^2$, $\frac{z}{\sqrt{r^2}+z^2}$ for $\cos \theta and \lambda dS for dq$ into the equation.

$$\begin{gathered} dE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{r^2+z^2}\left(\frac{z}{\sqrt{r^2+z^2}}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\frac{z\lambda dS}{{\left(r^2+z^2\right)}^{3/2}} \end{gathered}$$

Integrate above differential integral as,

$$\begin{gathered} E=\int dE \\ =\int \frac{1}{4\pi {\epsilon }_0}\frac{z\lambda dS}{{\left(r^s+z^2\right)}^{3/2}} \\ =\frac{\lambda }{4\pi {\epsilon }_0}\left(\frac{z}{{\left(r^s+z^2\right)}^{3/2}}\right)\left(2\pi r\right) \\ =\frac{\lambda }{2\pi \epsilon }\left(\frac{zr}{{\left(r^s+z^2\right)}^{3/2}\ne }\right) \end{gathered}$$

Thus, the electric field at a distance z above the center of a circular loop 

is$E=\frac{\lambda }{2{\epsilon }_0}\left(\frac{zr}{{\left(r^s+z^2\right)}^{3/2}\ne }\right)$.