Q4P - Introduction to Electrodynamics

Question

Find the electric field a distance z above the center of a square loop (side a) carrying uniform line charge A (Fig. 2.8). [Hint: Use the result of Ex. 2.2.]

                               

Step-by-Step Solution

Answer

The resultant electric field at a distance z above the center line of the square loop is $E=\frac{4a\lambda z}{4\pi {\epsilon }_0\left(z^2+{\displaystyle \frac{a^2}{2}}\right)\sqrt{z^2+\frac{a^2}{2}}}\hat{z}$

Step 1: Describe the given information

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.

Step 2: Define the coulombs law

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, 

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{qQ}}{{\mathbf{r}}^{\mathbf{2}}}$.

Step 3: Draw the diagram and show the components

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, 

$E^{\text{'}}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2\lambda L}{r\sqrt{r+L^2}}\right)$         ……. (1)

Here   $\lambda$is the line charge density,  $L$ is the length of the side,  ${\epsilon }_0$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

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

Step 4: Find the expression of electric field due to a side of loop

Theelectric field at the point, lying on the center of the square loop at a distance on 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)

$$\begin{gathered} E^{\text{'}}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2\lambda \left({\displaystyle \frac{a}{2}}\right)}{\left(\sqrt{z^2+{\displaystyle \frac{a^2}{4}}}\right)\sqrt{{\left(z^2+\frac{a^2}{4}\right)}^2+{\left({\displaystyle \frac{a}{2}}\right)}^2}}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{a\lambda }{\left(\sqrt{z^2+{\displaystyle \frac{a^2}{4}}}\right)\sqrt{z^2+\frac{a^2}{4}+\frac{a^2}{4}}}\right) \\ =\frac{a\lambda }{\frac{1}{4\pi {\epsilon }_0}\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}} \end{gathered}$$

Step 5: Simplify for resultant field

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

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

From the diagram, we can write

$\cos \theta =\frac{z}{r}$

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

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

Substitute $\frac{z}{\sqrt{z^2+\frac{a^2}{4}}} \mathrm{for} cos\theta , \frac{a\lambda }{4\pi {\epsilon }_0\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}} \mathrm{for} E^{\text{'}}$ into equation (2)

$$\begin{gathered} E=4\left(\frac{a\lambda }{4\pi {\epsilon }_0\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}}\right)\left(\frac{z}{\sqrt{z^2+\frac{a^2}{2}}}\right) \\ =\frac{4a\lambda z}{4\pi {\epsilon }_0\left(z^2+\frac{a^2}{2}\right)\sqrt{z^2+\frac{a^2}{4}}} \end{gathered}$$

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