# Do these steps demonstrate that acceleration of charged particle is proportional to current?

One formulation of Maxwell's Gauss Law for electric field is:

$$\bigtriangledown E = 4 \pi k \rho$$

This can be worked into the Divergence Theorem as follows:

$$\int\int_{A} F_\perp \:dA= 4\pi k \int\int\int_V \rho\:dV$$

As far as I can tell, the "outward pointing area normal" $F_\perp$ is the same as the current density or flux $J$ magnitude. I.e., $$F_\perp=J\cdot n=\frac{d}{dA}\left(\frac{dq}{dt}\right) \cdot n$$

$$\int\int_{A} J \cdot n \:dA\sim 4\pi k \int\int\int_V \rho\:dV$$

Now, if we assume the vector field is isotropic (flowing across a spherical surface), then $J \cdot n$ is constant across the surface, and so the LHS integral evaluates to the surface area of the sphere times the constant $J\cdot n$:

$$\int\int_{A} J \cdot n \:dA=J \cdot n \int\int_{A} \:dA=J\cdot n 4\pi r^2$$

where $r$ is the radius of the sphere. And then we get:

$$J\cdot n 4 \pi r^2\sim 4\pi k \int\int\int_V \rho\:dV$$

$$J \cdot n 4 \pi r^2\sim 4\pi k E$$

$$J \cdot n \sim \frac{kE}{r^2}$$

This already seems suspect since I believe by Coloumb's Law the RHS is proportional to the acceleration of a charged particle in the field. So this equation is tantamount to equating current density to acceleration. I do not have enough background in electromagnetism to know if such a statement makes sense or not.

But then also consider that by the conservation of charge we know that (continuity equation):

$$\bigtriangledown J = -\dot{\rho}$$

Substituing in the result obtained from the Divergence Theorem ($J\cdot n\sim \frac{kE}{r^2}$) gives:

$$\bigtriangledown \frac{k E}{r^2} \cdot n = -\dot{\rho}$$

Working out $\bigtriangledown \frac{k E}{r^2} \cdot n$ gives a mess that doesn't seem related to $\dot{\rho}$. Please help show how it is true or where I have gone wrong in my derivation.

• You seem to be very confused. What is $F_⊥$? Normal component of the electric field $\mathbf E$? And why it should be equal to normal component of the electric current density $\mathbf j$? – Ján Lalinský Dec 27 '13 at 10:13
• Note also that the current density is a vector while your $F_\perp$ is a scalar; the two should not be equivalent as they are different mathematical objects. – Kyle Kanos Dec 27 '13 at 16:32
• @KyleKanos Yes I meant the magnitude of $J$. I tried to fix this through an edit. – ben Dec 27 '13 at 18:39
• @JánLalinský Yes, generally for any field $F$, $F_\perp=F\cdot n$, for example as explained (en.wikipedia.org/wiki/Divergence_theorem). Since the charge density is on the RHS of the divergence theorem, then I suppose the $F$ we are talking about is $E$. – ben Dec 27 '13 at 18:48