Question

Show that Gauss’s theorem is consistent with Coulomb’s law. Using it, derive an expression for the electric field due to a uniformly charged thin spherical shell of radius r at a point at a distance y from the centre of the shell such that:
(I) y>r, and
(II) y<r.

Answer 1

Gauss’s Theorem

Gauss’s Law establishes that the total electric flux through any closed surface is equivalent to the enclosed net charge divided by the permittivity of free space:

\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{in}}}{\varepsilon_0} \]

Consistency with Coulomb’s Law

Consider a point charge \( q \) situated at the center of a spherical Gaussian surface with radius \( r \).

Due to the inherent symmetry, the electric field \( E \) possesses a constant magnitude and is directed radially outward across the surface:

\[ \oint \vec{E} \cdot d\vec{A} = E \oint dA = E \cdot 4\pi r^2 \] \[ \Rightarrow E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0} \Rightarrow E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r^2} \]

This equation precisely matches the expression for the electric field generated by a point charge as derived from Coulomb’s law, thus confirming the consistency of Gauss’s law.

Electric Field Due to a Uniformly Charged Spherical Shell

Let the total charge on the shell be \( q \), the radius of the shell be \( r \), and the observation point be at a distance \( y \) from the center.

Case I: \( y > r \) (Outside the shell)

Employ a spherical Gaussian surface with radius \( y \). The enclosed charge is \( q \).

\[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi y^2 = \frac{q}{\varepsilon_0} \Rightarrow E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{y^2} \]

This result indicates that the spherical shell effectively behaves as if its entire charge were concentrated at its center.

Case II: \( y < r \) (Inside the shell)

In this scenario, the Gaussian surface is positioned within the shell. The enclosed charge is 0.

\[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi y^2 = 0 \Rightarrow E = 0 \]

Therefore, no electric field is present inside a uniformly charged spherical shell.

Summary:

The electric field \( E \) produced by a uniformly charged spherical shell is described by the following expression:

\[ E = \begin{cases} \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{y^2}, & \text{for } y > r \\ 0, & \text{for } y < r \end{cases} \]