The Energy of a Continuous Charge Distribution (Griffiths EM Sect. 2.4.3 3rd ed)

[zackiechan]

Homework Statement

just confused about the derivation/justification

Relevant Equations

for point charge distribution $\frac{1}{2}\sum_{i=1}^{n}*q_i*V(\vec{r_i})$

for a continuous (volume, but the same $\frac{1}{2}$ applies to the line and surface formulas): $\frac{1}{2}*\int\rho V d\tau$

I'm working through Griffiths EM 3rd ed. in section 2.4.2 (point charge distribution) and 2.4.3 (continuous charge distribution).

I understand from the section on point charge distributions that when we add up all the work (excluding the work necessary in creating the charge itself), one clever way to do it is to sum and double-count every pairwise interaction (of course excluding self-interaction):
$$W = \frac{1}{2}\sum_{i=1}^{n}q_i\left(\sum_{j=1\\ j \neq i}^{n}\frac{1}{4\pi\epsilon_o}\frac{q_j}{\mathcal{r_{ij}}}\right) \hspace{10 mm} \mathcal{eqn. 1}$$
then correct for the double-counting by that factor of $\frac{1}{2}$ in front of the sum.

He goes on to recognize the term in parentheses as the potential at $V(\vec{r_i})$ and cleans it up as $$W = \frac{1}{2}\sum_{i=1}^{n}q_iV(\vec{r_i}). \hspace{10 mm} \mathcal{eqn. 2}$$

In the next section he just writes down the continuous volume charge density analog as: $$W = \frac{1}{2}\int\rho V d\tau \hspace{10 mm} \mathcal{eqn. 3}$$

He quickly mentions the line and surface analogs as $\frac{1}{2}\int\lambda V dl \hspace{5 mm} \frac{1}{2}\int\sigma V da$.

I'm confused as to why the $\frac{1}{2}$ is still in the integral for the continuous charge distribution cases?
When do we double count to have to correct by that factor of 2?
Is the $\frac{1}{2}$ a consequence of having recognized a piece of the double sum as a potential $V(\vec{\mathcal{r_i}})$?
Does using the potential automatically double-count?

 

[PeroK]

An integral (of a continuous distribution) is the limit of a sum of point charges (more and more, but smaller and smaller charges), so the factor of $1/2$ will remain through the limiting process.

 

[Abhishek11235]

To add above,we know that integration is limiting process. Here we just consider an infinitesimal charge $dq$. Now, all the mathematical steps are same as you did for point charge $q$. We will replace this with an infinitesimal charge $dq$. As $dq \rightarrow 0$ , we replace our sum with integration( This is due to Bliss Theorem)