Answer:

Gauss's Law relates the electric flux through a closed surface to the charge enclosed within that surface. For a uniformly charged thin spherical shell, we can use Gauss's Law to derive the expression for the electric field intensity (\(E\)) at a point outside the shell.

Consider a thin spherical shell of radius \(R\) with a uniform surface charge density (\(\sigma\)).

1. **Choose a Gaussian Surface:**

- Select a spherical Gaussian surface of radius \(r\) (where \(r > R\)) centered at the same point as the spherical shell. The Gaussian surface is a sphere of radius \(r\) that encloses the charged shell.

2. **Define the Electric Field:**

- The electric field (\(E\)) at any point on the Gaussian surface is radial, pointing outward, and its magnitude is the same at all points on the surface.

3. **Apply Gauss's Law:**

- Gauss's Law states: \(\oint \vec{E} \cdot \vec{dA} = \frac{Q_{\text{enc}}}{\varepsilon_0}\), where \(Q_{\text{enc}}\) is the charge enclosed by the Gaussian surface, \(\vec{E}\) is the electric field, \(\vec{dA}\) is the differential area vector, and \(\varepsilon_0\) is the permittivity of free space.

- For a thin spherical shell, the charge is distributed only on the outer surface, so \(Q_{\text{enc}} = \sigma \cdot 4\pi R^2\).

- The electric field (\(E\)) is constant in magnitude over the Gaussian surface, so the dot product \(\vec{E} \cdot \vec{dA}\) simplifies to \(E \cdot dA\).

- The differential area (\(dA\)) on the spherical Gaussian surface is \(4\pi r^2\).

- Applying Gauss's Law: \(E \cdot 4\pi r^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}\).

4. **Solve for Electric Field (\(E\)):**

- Cancel out common terms and solve for \(E\): \(E = \frac{\sigma R^2}{\varepsilon_0 r^2}\).

Therefore, the expression for the electric field intensity (\(E\)) at a point outside a uniformly charged thin spherical shell of radius \(R\) and surface charge density \(\sigma\) is given by \(E = \frac{\sigma R^2}{\varepsilon_0 r^2}\).

Verified answer

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Gauss's theorem relates the electric flux through a closed surface to the charge enclosed within that surface. In the case of a uniformly charged thin spherical shell, the electric field outside the shell is zero due to the symmetry of the charge distribution. The electric field inside the shell is also zero because there is no charge enclosed within a Gaussian surface placed inside the shell. Therefore, the expression for the electric field intensity at a point outside the shell would be zero.

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