Question

A thin spherical shell of radius \( R \) has a uniform surface charge density \( \rho \). Using Gauss' law, deduce an expression for the electric field:
(i) Outside the shell.
(ii) Inside the shell.

Hint:

Using Gauss' law, the electric field inside a spherical shell is zero, while outside it behaves as if the entire charge were concentrated at the center.

Answer 1

Step 1: Consider a thin spherical shell with a uniform surface charge density \( \rho \) and radius \( R \). The total charge \( Q \) on the shell is given by:
\[ Q = \rho \cdot 4\pi R^2 \] Step 2: Using Gauss' law, we consider a spherical Gaussian surface of radius \( r \), with \( r>R \) (outside the shell). The electric flux through the surface is given by:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] Since the charge is uniformly distributed on the shell, the electric field at any point outside the shell will be radially symmetric, and we can simplify the left-hand side as:
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \] Thus, the electric field outside the shell is:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \] Substituting \( Q = \rho \cdot 4\pi R^2 \), we get:
\[ E = \frac{\rho \cdot R^2}{\epsilon_0 r^2} \] Step 3: Now, consider the electric field inside the shell. For \( r>R \), by Gauss' law, the charge enclosed by the Gaussian surface is zero, because there is no charge inside the shell. Therefore, the electric field inside the shell is:
\[ E = 0 \] Conclusion: - (i) The electric field outside the shell is \( E = \frac{\rho \cdot R^2}{\epsilon_0 r^2} \)
- (ii) The electric field inside the shell is \( E = 0 \)