Step 1: Review of the Problem Setup.}
The problem involves the electric field inside a uniformly charged spherical shell. We know that the electric field produced by a charge distribution is determined by Gauss’s law. The spherical symmetry of the shell implies that the electric field at any point inside the shell should only depend on the geometry of the shell, not the distribution of charge itself.
Step 2: Applying Gauss’s Law.}
Gauss’s law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is written as:
\[
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}
\]
where \( Q_{\text{enclosed}} \) is the charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.
Step 3: Choosing a Gaussian Surface Inside the Shell.}
Now, consider a spherical Gaussian surface inside the shell. By symmetry, the electric field should point radially outward and be uniform over the spherical surface. Since the shell is uniformly charged, the charge enclosed by the Gaussian surface is zero. The reason is that there is no charge enclosed within the spherical shell (the charge resides only on the shell, not inside it).
Step 4: Consequence of Gauss’s Law.}
Since \( Q_{\text{enclosed}} = 0 \), from Gauss’s law, the total electric flux through the Gaussian surface must also be zero. The electric flux is the integral of the electric field over the surface area, and since the flux is zero, the electric field at any point inside the shell must also be zero.
Step 5: Final Conclusion.}
Therefore, the electric field inside a uniformly charged spherical shell is zero. This result holds for any point inside the shell, no matter how close or far it is to the center.