Question: Suppose you know that the electric field in distance $r > R$ from the center of a charged sphere with charge $Q$ and radius $R$ is given by:

$$
E = \frac{1}{4\pi\epsilon\_0} \frac{Q}{r^2}
$$

(Coulomb's law). 

From this I want to derive the field of electric field of a charged conducting plate with given charge density $\sigma := \frac{Q}{A}$ with the following heuristic argument: 

I start from the following three remarks:

* Suppose $r \approx R$
* The sphere is locally flat
* The field of the sphere is locally approximately homogenous

Then this local approximately flat small part the sphere should be the same as just a conducting plate.

Since the total area of the sphere is $A = 4\pi R^2$, I can conclude:

$$
E = \frac{Q}{4\pi r^2} \frac{1}{\epsilon\_0} = \frac{\sigma}{\epsilon\_0}
$$

should be the electric field of a conducting sphere (if you are close enough to it and not at the boundaries of the plate). 

However the correct formula is: 

$$
E = \frac{\sigma}{2\epsilon\_0}
$$

Or from another point of view: The formula which I derived is correct for two parallel conducting plates with equal but opposite charge densities.

So what is wrong with my argument that I get a factor $2$ wrong? Is there any way to modify this argument to get the correct result in a clear and convincing way? 

**Edit** 
As Phoenix87 commented: The difference of our local situation to to that of a plate is, that on the "local plate" (piece of the sphere) the field is zero on one side. If you have just a plate, you have an equal but opposite field on *both* sides. So the situation seems to be locally that of one plate of a parallel plate capacitor, then the formula above would be correct. But I think this argument is not very clear. Any ideas to improve it?

Answer: 
If you are sampling a [canonical ensemble](https://en.wikipedia.org/wiki/Canonical_ensemble) energy is not generally conserved. The detailed balance condition in such a setting reads
$$ \tag{1} e^{-\beta \mathcal H\_i} P(i \rightarrow j) = e^{-\beta \mathcal H\_j} P(j \rightarrow i), $$
where $i,j$ label the states in the canonical ensemble at temperature $T=1/k\beta$ and $\mathcal{H}\_i$ is the energy of the $i$-th state.

The stationary probability resulting from **(1)** can be shown to be
$$ \tag{2} \Pi\_i = \frac{e^{-\beta \mathcal H\_i}}{\sum\_j e^{-\beta \mathcal H\_j}}, $$
and this is used for example when the Markov chain is used to calculate averages of the form
$$ \tag{3} \langle \mathcal{O} \rangle =
\frac{ \int \mathcal D \phi \,\, \mathcal O [\phi] e^{-\beta \mathcal{H}[\phi]}}
{\int \mathcal D \phi \,e^{-\beta \mathcal{H}[\phi]}}. $$
I am not sure I fully understand your question, because in the title you talk about Molecular Dynamics and then in the body you talk about Monte Carlo. Anyway, here is my answer:

Molecular Dynamics does *not* satisfy detailed balance.

In a NVE molecular dynamics simulation, the transition

$$(\mathbf r(t), \mathbf v(t)) \to (\mathbf r(t+\tau), \mathbf v(t+\tau))$$

is *deterministic*, meaning that given a state $i$ there is one and only one a state $j$ such that

$$P(i\to j) =1 $$

the probability to go from $i$ to any other state is $0$:

$$P(i\to j') =0 \ \ \forall j'\neq j$$

The same is true also in a NVT simulation, but in a more subtle way, because the presence of the thermostat introduces "noise" in the dynamics of the system. Anyway, since Molecular Dynamics is based on the *deterministic* equations of motions of Newton, it will never satisfy detailed balance.

Conversely, Monte Carlo codes usually satisfy detailed balance, but they don't necessarily conserve energy: a well-known example is the Metropolis algorithm.