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

**OR**

Derive an expression for capacitance of parallel plate capacitor with a dielectric slab betw

Question

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

**OR**

Derive an expression for capacitance of parallel plate capacitor with a dielectric slab between the plates.

Accepted Answer

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> [!TLDR]
> Applying Gauss's law with a spherical Gaussian surface: outside the shell the field behaves exactly as if all the charge were concentrated at the centre ($E \propto 1/r^2$); inside the (empty) shell, the enclosed charge is zero, so $E = 0$ everywhere inside.
>
> Consider a thin hollow spherical shell of radius $R$, carrying a total charge $Q$ uniformly distributed over its surface. By symmetry, the electric field at any point due to the shell must be radial and must have the same magnitude at all points equidistant from the centre. This symmetry lets us use a concentric spherical **Gaussian surface** of radius $r$ through the point where we want the field, and apply Gauss's law:
> $$\oint \vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0}$$
> Since $\vec{E}$ is radial and constant in magnitude over the Gaussian sphere, and $d\vec{A}$ is also radial, $\vec{E}\cdot d\vec{A} = E\,dA$, so the left side becomes $E 	imes 4\pi r^2$ (the surface area of the Gaussian sphere).
>
> **(a) Point outside the shell ($r > R$):** The Gaussian sphere of radius $r$ encloses the entire charge $Q$ of the shell, so $q_{enclosed} = Q$. Gauss's law gives
> $$E(4\pi r^2) = \frac{Q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$
> This is identical to the field of a point charge $Q$ placed at the centre — i.e., a uniformly charged shell behaves, from outside, exactly like a point charge at its centre. …

Q.Using Gauss theorem derive an expression for electric field intensity due to uniformly charged hollow sphere (shell) at a point outside the shell and at a point inside the shell. OR Derive an expression for capacitance of parallel plate capacitor with a dielectric slab between the plates.

Concept understanding — Gauss Law

Why this formula?

Gauss's Law: Why It Holds

1. The Core Idea: Flux as "Flow" of Field

2. The Key Insight: Flux Depends Only on Enclosed Charge

3. Why Shape Doesn't Matter

4. Multiple Charges: Superposition

Q.Using Gauss theorem derive an expression for electric field intensity due to uniformly charged hollow sphere (shell) at a point outside the shell and at a point inside the shell. OR Derive an expression for capacitance of parallel plate capacitor with a dielectric slab between the plates.

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