91影视

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

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,$E=\frac{1}{4蟺{蔚}_o}\frac{qz}{{(z^2+R^2)}^{3/2}}$

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

From equation (i), we can write,

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

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.

From equation (i), we can write,

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

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

Hence, the field at this point is zero.

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:

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

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

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

$\begin{array}{c}{E}_{\max }=\frac{1}{{\text{4蟺蔚}}_{\text{o}}}\frac{4脳{10}^{鈭�6}\text{鈥塁}脳0.707脳0.02\text{鈥尘}}{{({(0.707脳0.02\text{鈥尘})}^2+{\left(0.02\text{鈥尘}\right)}^2)}^{\frac{3}{2}}}\\ =3.46x{10}^7\text{鈥塏/颁}\end{array}$

Hence, the value of the electric field is $=3.46x{10}^7\text{鈥塏/颁}$.