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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/4It is given that,the charge stored in the shell is \( \mathrm{{q}={16}~\mu{C}} \)the inner radius of the shell is \( \mathrm{{a}={4}~{c}{m}={0.04}~{m}} \)the outer radius of the shell is \( \mathrm{{b}={8}~{c}{m}={0.08}~{m}} \)It is required to find the magnitude of the electric field at the radial distance of,a) \( \mathrm{{r}={1}~{c}{m}} \)b) \( \mathrm{{r}={5}~{c}{m}} \)c) \( \mathrm{{r}={12}~{c}{m}} \)Explanation:To use Gauss's law to find the electric field at a distance \( \mathrm{{r}} \) from the center of the spherical shell, we need to choose a Gaussian surface that encloses the charge distribution. Since the charge is uniformly distributed throughout the volume of the shell, a natural choice is a spherical Gaussian surface with radius \( \mathrm{{r}} \).We can then apply Gauss's law, which states that the electric flux through any closed surface is proportional to the charge enclosed by the surface:\( \mathrm{\oint{\mathbf{{{E}}}}\cdot{d}{\mathbf{{{A}}}}={\frac{{{Q}_{{{e}{n}{c}}}}}{{\epsilon_{{0}}}}}~~~~~~~~~~~~~{\left({1}\right)}} \)where \( \mathrm{{\mathbf{{{E}}}}} \) is the electric field, \( \mathrm{{d}{\mathbf{{{A}}}}} \) is an infinitesimal area element on the Gaussian surface, \( \mathrm{{Q}_{{{e}{n}{c}}}} \) is the charge enclosed by the surface, and \( \mathrm{\epsilon_{{0}}} \) is the permittivity of free space.Since the charge distribution is spherically symmetric, the electric field must also be spherically symmetric. Therefore, we can choose the Gaussian surface to be a sphere of radius \( \mathrm{{r}} \), centered at the center of the shell. The electric f ... See the full answer