tempneff said:
I am in vector calculus as well, but is it really necessary to derive them that way? Basic calculus is the only prerequisite for the course. Does anyone know how to derive them from ∫E ds
That would be vector calculus. Gauss' Law is
\nabla \cdot \mathbf{D} = \rho
Taking the integral over volumetric space and using the divergence theorem,
\int \mathbf{D} \cdot d \mathbf{S} = \int \rho dV = Q_{enclosed}
If we assume a homogeneous medium then finally,
\int \mathbf{E} \cdot d \mathbf{S} = \frac{Q_{enclosed}}{\epsilon}
Now in electrostatics, Maxwell's Equations state that the curl of the electric field is zero. That is,
\nabla \times \mathbf{E} = 0
This allows us to represent the electric field as the gradient of a scalar since the curl of a gradient is always zero. Thus, we choose this scalar to be the electric potential.
\mathbf{E} = -\nabla V
If we take the line integral of the electric field from some point B to A we get via the gradient theorem,
\int_b^a \mathbf{E} \cdot d\mathbf{\ell} = V_b - V_a
which is path independent because the electrostatic field is conservative (by virtue of being curl free). Finally, we can use Gauss' Law to see that
\nabla^2 V = -\frac{\rho}{\epsilon}
which is Poisson's Equation.
