Question

What do the Gauss's Law in differential and integral form mean? Why are they different from the normal Gauss's law?

Answer 1

They are same , just stated differently

Answer 2

I meant, why are they stated differently? What do the different forms signify?

Answer 3

\[\nabla \cdot E = \rho/\epsilon \]

take integrals of both sides

\[\int \int \int \nabla \cdot E dv = \int \int \int \rho/\epsilon dv \]

according the divergence theorem , right sides is equal to surface area.

rho is charge per volume , when you integrate w.rt. to volume you get charge Q

\[\oint E da=\frac{Q}{\epsilon }\]

Answer 4

differential form is preferred because since electric is vector, vector operation like divergence are more convience

Answer 5

The integral representations of Maxwell's Equations are global statements (they relate quantities that are summed over surfaces) that allow us to calculate things like the flux through some generic surface. The differential representations are local statements (they relate quantities at a given point in space), and are often times more mathematically useful for advanced calculations.

Answer 6

@imranmeah91: I already knew about the theorems and derivations, however your clarifications helped me understand better. Thanks a lot.
@jemurray3: thanks for your additional tips on Maxwell's equations and the use of different forms.