Question

Using Gauss' theorem, derive an expression for electric field intensity due to a uniformly charged hollow sphere (shell) at a point outside the shell and at a point inside the shell.

Hint:

The electric field outside a uniformly charged hollow sphere behaves as if the charge were concentrated at the center. Inside the sphere, the electric field is zero.

Answer 1

Step 1: Electric field outside the hollow sphere.
We apply Gauss' law to calculate the electric field at a point outside the hollow sphere. According to Gauss' law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] where \( \oint \vec{E} \cdot d\vec{A} \) is the electric flux through a closed surface, \( Q_{\text{enc}} \) is the charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.
For a point outside the hollow sphere, we choose a spherical Gaussian surface with radius \( r \) (where \( r \) is greater than the radius of the shell). The charge enclosed by this surface is the total charge \( Q \) on the shell. Because of the symmetry of the situation, the electric field \( E \) is radially symmetric and has the same magnitude at every point on the Gaussian surface. Therefore, the flux is: \[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \] Solving for \( E \), we get: \[ E = \frac{Q}{4\pi \epsilon_0 r^2} \] Thus, the electric field outside the hollow sphere behaves as if the entire charge \( Q \) were concentrated at the center of the sphere, similar to the electric field due to a point charge.
Step 2: Electric field inside the hollow sphere.
Now, consider the electric field at a point inside the hollow sphere. Since there is no charge inside the spherical shell (the charge is distributed on the outer surface), the charge enclosed by a Gaussian surface inside the shell is zero. According to Gauss' law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} = 0 \] Thus, the electric field inside the hollow sphere is zero. This result is a consequence of the symmetry of the situation, where the charges on the shell produce no electric field at points inside the shell.
Step 3: Conclusion.
Thus, the electric field intensity outside the shell is: \[ E = \frac{Q}{4\pi \epsilon_0 r^2} \] and the electric field inside the shell is: \[ E = 0 \]