Question

Find out the formula for the electric field at a point outside a uniformly charged spherical shell by using Gauss's theorem.

Hint:

Take a concentric spherical Gaussian surface of radius r > R; flux \(=E\,(4\pi r^2)=q/\varepsilon_0\) gives \(E=q/(4\pi\varepsilon_0 r^2)\).

Answer 1

Step 1: State Gauss's theorem.
Gauss's theorem says the total electric flux through any closed surface equals the charge enclosed divided by \(\varepsilon_0\):
\[ \oint \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0} \]
Step 2: Describe the setup and choose a Gaussian surface.
Let a spherical shell of radius \(R\) carry a total charge \(q\) spread uniformly over it. To find the field at an external point P at distance \(r\) from the centre (\(r > R\)), imagine a concentric spherical Gaussian surface of radius \(r\) passing through P.

Step 3: Use symmetry.
By spherical symmetry, \(\vec{E}\) has the same magnitude everywhere on this surface and points radially outward, parallel to \(d\vec{S}\) at every point. Hence \(\vec{E}\cdot d\vec{S} = E\,dS\).

Step 4: Evaluate the flux.
\[ \oint \vec{E}\cdot d\vec{S} = E\oint dS = E\,(4\pi r^2) \]since the area of the Gaussian sphere is \(4\pi r^2\).

Step 5: Find the enclosed charge.
The Gaussian surface (radius \(r > R\)) encloses the whole shell, so \(q_{enc} = q\).

Step 6: Apply Gauss's theorem and solve for E.
\[ E\,(4\pi r^2) = \frac{q}{\varepsilon_0} \]\[ E = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r^2} \]
Step 7: Interpret.
The field outside is exactly the same as if the entire charge \(q\) were concentrated at the centre of the shell.
\[\boxed{\ E = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q}{r^2}\quad (r > R)\ }\]