91Ó°ÊÓ

Figure 23-22 shows, in cross section, three solid cylinders, each of length L and uniform charge Q with a Gaussian surface of equal radius.

The Gaussian surface is known as a closed surface in three-dimensional space such that the flux of a vector field is calculated. Thus, considering the flux concept, to get the electric field through the Gaussian surface of the same radii yields an equal electric field that only depends on the charge value inside the surface.

Formula:

The electric flux through any closed surface is,

${\mathrm{Ï•}}_{\mathrm{E}}=âˆ\circledR \stackrel{→}{\mathrm{E}}.\stackrel{→}{\mathrm{dA}}$

${\mathrm{Ï•}}_{\mathrm{E}}=\frac{{\mathrm{q}}_{\mathrm{enclosed}}}{{\mathrm{Î\mu }}_0}$ ….. (i)

Using the given figure, you can get that the area vector is parallel to the electric field at any point on the surface, thus the flux value using equation (i) can be given as:

$\begin{array}{c}{\mathrm{Ï•}}_{\mathrm{E}}=âˆ\circledR \mathrm{EdAcos}0°\\ =\mathrm{EA}\end{array}$

Again, using equation (i), the electric field at a point within any Gaussian surface can be given as:

$\begin{array}{c}\mathrm{EA}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{Î\mu }}_0}\\ \mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{´¡Î\mu }}_0}\end{array}$

But, as given that the Gaussian radius for all three cases is the same. Thus, they have the same area.

Now, the charge enclosed within each cylinder is $\mathrm{Q}$ irrespective of its radius.

Thus, the value of the electric fields though all the Gaussian area with the given cylinders is same that is given by:

$\mathrm{E}=\frac{\mathrm{Q}}{{\mathrm{´¡Î\mu }}_0}$

Hence, the rank of the surfaces according to their electric fields is $\left(\mathrm{a}\right)=\left(\mathrm{b}\right)=\left(\mathrm{c}\right)$.