Question

State and prove Gauss's theorem in electrostatics. Calculate the electric field intensity at a point outside a hollow uniformly charged sphere.

Hint:

Gauss's theorem: \(\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}\). Electric field outside a uniformly charged sphere: \(E = \frac{Q_{\text{enc}}}{4 \pi \epsilon_0 r^2}\).

Answer 1

Gauss's Theorem states that the total electric flux (\(\Phi_E\)) passing through any closed surface is proportional to the total charge enclosed within the surface. Mathematically, it is expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] Where: - \(\oint \vec{E} \cdot d\vec{A}\) is the surface integral of the electric field \(\vec{E}\) over the closed surface \(A\),
- \(Q_{\text{enc}}\) is the total charge enclosed within the surface,
- \(\epsilon_0\) is the permittivity of free space.
Proof:
The flux through a closed surface is the integral of the electric field over the surface area. If the charge distribution is symmetric, such as in the case of spherical symmetry, the electric field is normal to the surface at all points, and the magnitude is constant. Thus, Gauss's law simplifies to: \[ \Phi_E = E \cdot A \] Where \(E\) is the magnitude of the electric field and \(A\) is the surface area. Using Gauss's law, we can calculate the electric field at any point outside a spherical charge distribution.
Electric Field Outside a Hollow Uniformly Charged Sphere:
For a uniformly charged spherical shell, we apply Gauss's theorem to find the electric field outside the sphere. Consider a spherical Gaussian surface of radius \(r\) greater than the radius of the sphere. Since the charge is uniformly distributed, by symmetry, the electric field at any point on the surface will be radially outward and have the same magnitude at all points.
By Gauss's law, we have: \[ \Phi_E = E \cdot A = \frac{Q_{\text{enc}}}{\epsilon_0} \] Where \(A = 4\pi r^2\) is the surface area of the spherical Gaussian surface and \(Q_{\text{enc}}\) is the total charge on the sphere. Therefore, the electric field intensity \(E\) is given by: \[ E = \frac{Q_{\text{enc}}}{4 \pi \epsilon_0 r^2} \] Thus, the electric field outside the hollow uniformly charged sphere behaves exactly as if all the charge were concentrated at the center of the sphere.