Question

Figure 23-23 shows, in cross section, a central metal ball, two spherical metal shells, and three spherical Gaussian surfaces of radii R, 2R, and 3R, all with the same center. The uniform charges on the three objects are: ball, Q; smaller shell, 3Q; larger shell, 5Q. Rank the Gaussian surfaces according to the magnitude of the electric field at any point on the surface, greatest first.

Short answer

The rank of the Gaussian surfaces according to the magnitude of the electric field at any point on the surface is${\mathrm{E}}_1={\mathrm{E}}_2={\mathrm{E}}_3$.

Step-by-step solution

The given data: 

Figure 23-23 shows, the central metal ball with charge $\mathrm{Q}$ and two spherical metal shells ($3Q$ and $5Q$), and three Gaussian surfaces of radii $\mathrm{R}$, $2R$, and $3\mathrm{R}$ are given.

Understanding the concept of Gaussian surface:

Here, the total charge within a Gaussian charge with its given area is used to calculate the electric field at any point of the Gaussian surface.

Formula:

The electric field at any point of a Gaussian surface,

$\mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{A\epsilon }}_0}$ ….. (i)

Here, $\mathrm{E}$ is the electric field, ${\mathrm{q}}_{\mathrm{enc}}$ is the enclosed charge, $A$ is the area, and ${\epsilon }_0$ is the permittivity of the free space.

Calculation of the rank of the Gaussian surfaces at any point:

Now, using the data from the figure, the charge enclosed by Gaussian surface 1 can be given as:

${\mathrm{q}}_{\mathrm{enc}}=\mathrm{Q}$

Thus, using the above charge value in equation (i), the electric field at any point of Gaussian surface 1 of radius$\mathrm{R}$can be given as:

${\mathrm{E}}_1=\frac{\mathrm{Q}}{4{\mathrm{\pi R}}^2{\mathrm{\epsilon }}_0}(\because \text{Surfaceareaofasphere,}\mathrm{A}=4{\mathrm{\pi R}}^2)$

Now, using the data from the figure, the charge enclosed by Gaussian surface 2 can be given as:

$\begin{array}{c}{\mathrm{q}}_{\mathrm{enc}}=\mathrm{Q}+3\mathrm{Q}\\ =4\mathrm{Q}\end{array}$

Thus, using the above charge value in equation (i), the electric field at any point of Gaussian surface 2 of radius$2R$can be given as:

$\begin{array}{c}{\mathrm{E}}_2=\frac{4\mathrm{Q}}{4\mathrm{\pi }{\left(2\mathrm{R}\right)}^2{\in }_0}(\because \text{Surfaceareaofasphere,}\mathrm{A}=4{\mathrm{\pi R}}^2)\\ =\frac{\mathrm{Q}}{4{\mathrm{\pi R}}^2{\in }_0}\end{array}$

Now, using the data from the figure, the charge enclosed by Gaussian surface 3 can be given as:

$\begin{array}{c}{\mathrm{q}}_{\mathrm{enc}}=\mathrm{Q}+3\mathrm{Q}+5\mathrm{Q}\\ =9\mathrm{Q}\end{array}$

Thus, using the above charge value in equation (i), the electric field at any point of Gaussian surface 3 of radius$3R$can be given as:

$\begin{array}{c}{\mathrm{E}}_3=\frac{9\mathrm{Q}}{4\mathrm{\pi }{\left(3\mathrm{R}\right)}^2{\in }_0}(\because \text{Surfaceareaofasphere,}\mathrm{A}=4{\mathrm{\pi R}}^2)\\ =\frac{\mathrm{Q}}{4{\mathrm{\pi R}}^2{\in }_0}\end{array}$

Hence, the rank of the rank of the Gaussian surfaces is ${\mathrm{E}}_1={\mathrm{E}}_2={\mathrm{E}}_3$.