One E field eq in/outside sphere of charge?

[cefarix]

What is the single equation that describes the E field both inside, outside, and at the surface of a sphere of charge? At the hyperphysics website they give two different equations for both situations, the outside eq is inverse square, and the inside equation is direct linear.

 

[Doc Al]

What do you mean by "single equation"? As hyperphysics correctly states, the field within the surface depends linearly on the distance from the center, but outside the surface it depends on the distance squared. That is the formula describing the field at any point.

 

[cefarix]

The equations are two though, one for the field inside, and separate one for the field outside. I suppose there is no way to directly combine these two equations. However can't we derive an equation which, I guess would be in integral form, summing up the flux through infinitesimal patches over the sphere's surface?

 

[quasar987]

The one equation you appear to be refeering to is the first maxwell equation

$$\vec{\nabla}\cdot \vec{E} = \frac{\rho}{\epsilon_0}$$

or

$$\int \vec{E}\cdot d\vec{a} = \frac{Q_{int}}{\epsilon_0}$$

in integral form.

When we solve it for a charged sphere though, it turns out that E is of a different form inside and outside and is undefined directly at the surface.

 

[krab]

The equations are two though, one for the field inside, and separate one for the field outside. I suppose there is no way to directly combine these two equations. However can't we derive an equation which, I guess would be in integral form, summing up the flux through infinitesimal patches over the sphere's surface?

There is just one equation. However, because the gradient is discontinuous across the surface, when we write the equation as an analytic function, it must be split into two regimes. If this bothers you, realize that if you were to try to find a single continuous function to describe the density rho both inside and outside the sphere, you would end up with the same problem: no way to do it.