What is the analog between charges and mass with regards to vector fields?

Question

A way to think about gravitational fields is that they spawn as a consequence of mass literally bending time and space.

Is there an analog of this concerning charges and the electric field intensity?

Answer 1

I don't know how analogous this really is, but charges push and pull electric fields from themselves. This is Gauss' law,

$$\Phi_E=\oint_{\partial\Omega}d^2\mathbf r\cdot\mathbf E=\int_\Omega d^3\mathbf r\frac{\rho}{\epsilon_0}=\frac{Q_{int}}{\epsilon_0},$$

which pictorically states that the number of electric field lines coming out of a surface (the electric flux $\Phi_E$) is proportional to the amount of charge inside it ($Q_{int}$). Positive charges "bend" the electric field away from them, while negative charges "bend" it towards them.

The usual picture used to illustrate this is the electric field lines between two equal and opposite charges placed a distance from each other in free space,

Note how the lines flow out of the positive charge and into the negative charge.

Answer 2

The short answer is no. The big thing that makes vector fields $\vec E$ and potentials $\Phi_E$ useful in electrodynamics is the superposition principle. If you have two charges, the total electric potential is the sum of the two electric potentials coming from each. Similarly if you have a continuum of charges, etc. Likewise, the vector fields $\vec E$ add. The superposition principle is the statement that electrodynamics is a linear theory.

Gravity is not linear. The metric $g_{\mu\nu}$ due to two point masses is not the sum of the two metrics coming from each.

If you work in the Newtonian approximation of gravity (which is linearized Einstein equation + all time derivatives vanish) -- i.e. if you work in the high-school physics approximation of gravity, which is great for many cases-- then certainly you have an equation

$$\vec\nabla^2\phi = 4\pi G \rho$$

where $\rho$ is the matter density and $\phi$ is the gravitational potential (which is related to the metric by $g_{00} \approx -1 + 2\phi$) and the derivative is in the spatial coordinates only. You can define a gravitational field $\vec \nabla \phi$ and treat this similar to electrostatics if you wish (but all forces are attractive). But this theory is not relativistic.