We liked these questions, playing around with the old ideas that were used by the founders, Coulomb, Gauss, Faraday.
The electic charge density on a small plane piece $dO$ of a charged conducting surface $S$ is called electric induction, because it can be induced by an electrostatic charged, small ball approaching the conducting surface.
$$D = 1/\varepsilon\ E = \frac{dQ}{dS} $$.
By elementary surface integrals over differential surface densities, the induction field $D$ is a map from areas between coordinate lines of constant coordinate values. On a sphere the areas are the wedge product of the euclidean metric differentials of the coordinates
$$ dS = r d\theta \wedge r \sin \theta d\phi$$
These maps have an orientation and are expressed as signed wegde products of the basic linear appoximations to the coordinate lines.
In principle, a differential form $dx$ counts, how many lines of constant $x=c$ any vector is crossing.
$$ dx (a e_x + b e_y) == a $$
The signed wedge product defines an antisymmetric map with two arguments
$$ dx \wedge dy (a e_x + b e_y, c e_x + d e_y) == a d - b c$$
Physicists prefer to call this the cross product of the two tangent vectors, perpendicular to the rectangle spanned by $dx, dy$ between coordinate lines.
Mathematically there is no norm nor a notion of orthogonality in multiple integrals, its a normless summing of a density function over elementary coordinate n-cubes with an orientation scheme, such that the integral over inner surfaces of a partion in n-cubes vanish.
So, under a surface integral,
$$D = \rho r^2 \sin \theta d\theta\wedge d\phi $$
simply counts the signed areas with, the sign given by the order and growing direction of the cordinate edges.
The electric induction field $D$ is a surface density 2-form and there is no need to make a vector of it.
The result is Coulomb's celebrated law, that connects the surface charge density 2-form with an acceleration 2-form acting on a massive point charge, in this case as arguments two velocity vectors, one in the directions of space and one in time.
$$\frac{charge}{length^2}\ \{ \frac{A s}{m^2} \}\ \to \
\frac{energy}{length} \ \{ \frac{A V s}{m} \} == \frac{mass \ length} {time^2} \ \{ \frac{kg m}{s^2} \}$$
The electric field definition includes a euclidean norm, namely the square of velocities (not lengths). So the electric field definition needs a metric. The metric is readily introduced by a metric on the linear maps
$$ dx_i . dx_k == (g^{-1})ik $$
The inverse of the metric is used, because Einstein introduced $g$ on vectors. In our case we chose ,what the euclidean metric imposes for the pure coordinate differentials
$$d\phi \cdot d\phi = 1/( r^2 \sin^2 \theta),\ \ d\theta \cdot d\theta = 1/ r^2$$
The system is orthogonal, the non-diagonal products are zero.
The so called Hodge product of forms yields the electric force 2-form by the metric product with the 4-volume form of space-time.
$$ \left( \ dct \wedge dr \wedge r\ d\theta \ \wedge r \sin \theta\ d\phi \ \right) \cdot \
\left(\ D(R) \frac{R^2}{r^2} \ r\ d\theta \wedge r\ \sin \theta \ d\phi \ \right) == \ D(R) \frac{R^2}{r^2} dct \ \wedge \ dr $$
By spherical symmetry, the $D$-Field outside the conducting sphere is
$$ D\left(r\right) = \ \frac{ Q} {(4 \pi R^2)}\ \ \left(\frac{R}{r}\right)^2 $$
no electric charge inside, please.
We know that the electic field $E = c/r^2$ in any coordinate system is smooth, so we have
$$\partial_\phi E_r = \partial_r E_\phi$$
etc, that implies rot E=0 and so it follows, $E$ is the gradient of a potential function up to a constant.
Let's set
$$ \Phi\left(r\right) \ = \ \int\ E \ dr \ = \ 0$$ at infinity.
Now with
$$\Phi\left(r\right) = \frac{Q}{4 \pi \varepsilon} \ \frac{1}{r}$$
independent of the sphere radius $R$. The continuity at the surface with the jump in the derivative is given by the charge density, the potential in the inner is constant (Faraday cage) completely independent of the charge distribution. In this picture, there is no force between charges on the symmetric charged sphere, all forces point radial to infinity.
But a charged point does not move. So what?
Yes, on the charged sphere, the force towards the center is removed for any charged point on the surface.
So, in this case, the particle is subject to its own radial outward force.
This is just the opposite of Griffiths idea.
Quite parallel to the mainstream vector analysis, for me, it was a fascinating area of thoughts that formed modern differential topology.
The names are Gauss, Maxwell, Hilbert, Poincare, Cartan and the whole bunch of modern field theorists, that 'ironed out' smooth fields of any local algebraic structure like vectors, tensors, n-forms, spinors, metrics, Clifford algebras or Lie-groups over any kind of manifolds, while academic teaching of electrodynamics was reduced to study and exercise those absurd formula collections for grad, div, rot and Laplacian in eleven coordinate systems.
Bottom line:
$$ A = \sum_{j=0}^3 A_j dx^j $$
is the gauge field modifying the momentum operator in canonical mechanics an in quantum theory.
The exterior derivative $d$ maps a $n$-form to a $n-1$ form by the wedge product of the derivative of the coefficent with the wegde product of the basis
$$ dA = \sum_{j}^3\ \left( \ \sum_{k=0}^3 \partial_k A_j dx^k \ \right) \ \wedge dx^j = \sum_{j,k=0}^3 F_{kj} dx^k \wedge dx^j $$
$$ F_{jk} == \partial_j A_k-\partial_k A_j$$
$$ E = dct \wedge \left( \sum_{i=1}^3 E_i dx^i\ \right) , \ *B = \sum_{ijk} B_i \ \varepsilon_{ijk}\ dx^i \wedge dx^k$$
is the electric force 2-form acting on velocities classically.
By the constancy of the linear approximation and antisymmetry we have
$$ ... dx^k\wedge .... \wedge dx^k ...==0 , \ \ d dx^k == 0$$
It follows
$$ dF = 0 \ \ \to \nabla \cdot E == 0 \ \ , \ \ \nabla \times B + \partial_{ct}\ E =0$$
This is the homogenous group of the Maxell equations.
The Hodge product with the 4-volume is called hodge dual and is denoted by an asterisk prefix. The exterior derivative of the charge-current generated fields $*F = D,H$ d yield the 3-form of currents
$$ d \star F \ \ == j $$
This is the inhomgenuous part of the Maxwell equations
$$ \nabla E = \rho \ \ , \ \ \nabla \times B + \partial_{ct} E = j $$
3-forms are surface integrand over a 4-volume cube.
Their Hodge duals are the component of the 4-vector of currents.
The time component is the charge density as a flow of charge in a volume in time direction.
The space components are the current components flowing over 2-surfaces of a 3-volume in a time interval
This in form language yields
$$ *F = S, \ \ dS = *j, \ \ d*j=0 $$
$d*j= dd*F = 0$ is conservation of 4-current and constant of total charge in an isolated system.
The current through spatial surfaces equals the change if charged in the 2-volume over a time interval.
$$\partial_ct \rho + \nabla \cdot j ==0 $$
Finally the dd - circle is closed by a ^(d^{-1} , the integral formula for the potential 1-form fro stationary systems of charges and currents,
The potential 1-form of mechanical momentum local gauge with respect to velocity is the convolution with the Greens function of
$$d*d = \partial_{ct,ct} - \Delta$$
namely, the Coulomb potential in empty space
$$ A = \int \frac{j(\xi)}{|x-\xi|}$$
best known for the charge, but by all these algebraic identities also valid for constant currents and their vector potential. The 1-form
$$A = \Phi dct + A_i dx^i \mapsto (E/c -e/c\Phi )dt + (p_i - e/c A_i) dx^i$$
By the canonical formalism, E,p induce translations in paramter space $t,x$ while the gauged momentum forms are the velocities of material transport.
Hope it helps
