Question

State and explain Gauss's law in electrostatics. Using it, find the electric field due to a uniformly charged thin spherical shell (charge = q and radius = R) at 

(A) external point of shell 

(B) internal point of shell 

(C) on the surface of shell. 
 

Hint:

Gauss's law simplifies the calculation of the electric field in situations with spherical symmetry, such as around a uniformly charged spherical shell.

Answer 1

Gauss Law: Gauss law states that the electric flux \( \Phi_E \) through any closed surface is proportional to the charge \( Q_{\text{enclosed}} \) enclosed by the surface. Mathematically: \[ \Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}, \]

where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, and \( \epsilon_0 \) is the permittivity of free space.

Now, applying Gauss's law to a uniformly charged thin spherical shell, we can find the electric field at various points:

(A) At an external point (at a distance \( r > R \) from the center of the shell): Since the charge distribution is spherically symmetric, we can use a spherical Gaussian surface of radius \( r \) (where \( r > R \)). The total charge \( Q \) is enclosed within this Gaussian surface, and the electric field at a distance \( r \) is given by:

\[ E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}, \]

so the electric field is:

\[ E = \frac{q}{4 \pi \epsilon_0 r^2}. \]

(B) At an internal point (at a distance \( r < R \) from the center of the shell): For a point inside the uniformly charged spherical shell, the electric flux through the Gaussian surface is zero because no charge is enclosed within the Gaussian surface. Thus, the electric field inside the shell is:

\[ E = 0. \]

(C) At a point on the surface of the shell (at \( r = R \)): At the surface, the electric field will have the same magnitude as at any external point on a spherical shell with radius \( r = R \). Thus, the electric field is:

\[ E = \frac{q}{4 \pi \epsilon_0 R^2}. \]