Q24P - Fundamentals Of Physics

Question

A thin non-conducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R (Fig. 22-48). The central perpendicular axis through the ring is a z axis, with the origin at the center of the ring. What is the magnitude of the electric field due to the rod at (a)  $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}$ and (b) $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{\infty }$ ? (c) In terms of R, at what positive value of z is that magnitude maximum? (d) If  $\mathit{R}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{2.00}\mathbf{ }\mathit{c}\mathit{m}$ and $\mathbf{Q}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{4}\mathbf{.00}\mathbf{ }\mathbf{\mu C}$, what is the maximum magnitude?

Step-by-Step Solution

Answer

1. The magnitude of the electric field due to rod at $z = 0$ is zero.
2. The magnitude of the electric field due to rod at $z=\infty$ is zero.
3. The value of z at which the magnitude of the field has maximum is $0.707R$.
4. The maximum magnitude for the given values is $3.46x{10}^7\text{\hspace{0.17em}N/C}$.

Step 1: The given data

• A thin non-conducting of charge $Q$  is bent into circle of radius, $R$.
• The given values: $R = 2.00\text{ cm}$ and  $Q = 4.00 \text{\hspace{0.17em}μC}$.

Step 2: Understanding the concept of electric field

Using the concept of the electric field of a dipole due to a ring, we can get the required value of the electric field and by differentiating it, we can get the required maximum magnitude of the electric field and its position.

 

Formula:

The electric field along the axis at the centre of a charged ring, $E=\frac{1}{4\pi {\epsilon }_o}\frac{qz}{{(z^2+R^2)}^{3/2}}$

Where,$R$  is radius of the ring, $z$  is distance of the point from the circle,  is charge of the particle.

Step 3: a) Calculation of the field at z = 0

From equation (i), we can write,

$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{qz}}{{({\mathrm{z}}^2+{\mathrm{R}}^2)}^{3/2}}$

From the figure 22-48, the center of the circle is at the same distance from every part of the coil. Therefore, using the symmetry and equation (i), we can say that the field vanishes at the center. It is because, when any part of the coil creates a field, an equal and opposite field is created by the opposite part of the coil.

Hence, the electric field at this point is zero.

Step 4: b) Calculation of the field at z   =   ∞

From equation (i), we can write,

$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{qz}}{{({\mathrm{z}}^2+{\mathrm{R}}^2)}^{3/2}}$

 

In the above equation, there is a term $1/{(z^2+R^2)}^{3/2}$, which will become zero as $z=\infty$. 

Hence, the field at this point is zero.

Step 5: c) Calculation of the value of z at which we get the maximum field

Differentiating equation (i) which is the equation of the electric dipole due to a circular ring, and setting it equal to zero (to obtain the location where it is maximum), we can get the value of z as follows:

$\begin{array}{c}\frac{\mathrm{d}}{\mathrm{dz}}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{qz}}{{({\mathrm{z}}^2+{\mathrm{R}}^2)}^{\frac{3}{2}}}\right)=0\\ \frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{{\mathrm{R}}^2-2{\mathrm{z}}^2}{{({\mathrm{z}}^2+{\mathrm{R}}^2)}^{\frac{5}{2}}}=0\\ \mathrm{z}= +\frac{\mathrm{R}}{\sqrt{2}}\\ =0.707\mathrm{R}\end{array}$

Hence, the value of the distance of the point where we get maximum electric field is $0.707R$.

Step 6: d) Calculation of the magnitude of the maximum field

Plugging the given values back into equation (i) and using above value, we can get the magnitude of the maximum electric field as follows:

$\begin{array}{c}{E}_{\max }=\frac{1}{{\text{4πε}}_{\text{o}}}\frac{4\times {10}^{-6}\text{\hspace{0.17em}C}\times 0.707\times 0.02\text{\hspace{0.17em}m}}{{({(0.707\times 0.02\text{\hspace{0.17em}m})}^2+{\left(0.02\text{\hspace{0.17em}m}\right)}^2)}^{\frac{3}{2}}}\\ =3.46x{10}^7\text{\hspace{0.17em}N/C}\end{array}$

Hence, the value of the electric field is $=3.46x{10}^7\text{\hspace{0.17em}N/C}$.