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 states that the total electric flux through a closed surface is equal to the net charge enclosed 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 \( q \) placed at the center of a spherical Gaussian surface of radius \( r \).

Due to symmetry, the electric field \( E \) is constant in magnitude and radially outward over 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 is exactly the expression for electric field due to a point charge from Coulomb’s law, hence Gauss’s law is consistent with it.

Electric Field Due to a Uniformly Charged Spherical Shell

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

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

Use a spherical Gaussian surface of radius \( y \). Enclosed charge = \( q \)

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

This means the shell behaves like a point charge concentrated at its center.

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

Now, the Gaussian surface lies inside the shell. Enclosed charge = 0

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

Hence, no electric field exists inside a uniformly charged spherical shell.

Summary:

The electric field \( E \) due to a uniformly charged spherical shell is given by:

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