Section 2.7: Gauss’s Law Introduction
Gauss’s Law relates the net electric flux through a closed surface to the total charge enclosed within that surface.
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \] where the integral is over a closed surface, \( Q_{enc} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space.
Gauss’s Law is particularly useful for problems with spherical, cylindrical, or planar symmetry.
Example 1
Use Gauss’s Law to derive the electric field outside a point charge \( q \).
Choose a spherical Gaussian surface of radius \( r \) centered on the charge.
\[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4 \pi r^2 \]
\[ E \cdot 4 \pi r^2 = \frac{q}{\epsilon_0} \quad \Rightarrow \quad E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \]
Thus, Coulomb’s law is recovered from Gauss’s Law.
Practice Problems
- State Gauss’s Law in words and explain its physical meaning.
- Apply Gauss’s Law to a uniformly charged spherical shell; describe the field inside and outside.
- Why is Gauss’s Law particularly useful for high-symmetry problems?
- Calculate the flux through a cube enclosing a point charge at its center.
- Does Gauss’s Law depend on the shape of the surface?