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Suppose there is a charge distribution in region V and zero charge density outside V such that electric field goes to zero infinitely far away. How to prove that this charged body does not exert an electrostatic force on itself? Here’s how far I’ve got: $$F=\int_V \rho \vec{E} dV=\int_V -\nabla (\Phi \rho) + \Phi \nabla \rho dV$$ Now the first part of the integral vanishes if we integrate beyond the boundary of the charged body, by stokes theorem and because there’s zero charge on the outside. $$\int_V \rho \vec{E} dV=\int_V \Phi \nabla \rho dV$$ I can’t do anything after this. Can you help me please?

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  • $\begingroup$ You have to use the fact that $\Phi$ is only due to $\rho$ of the system, not due to some external charges. $\endgroup$ – Ján Lalinský Jan 18 at 13:18
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$\Phi(\vec{r})\propto\int_V\frac{\rho(\vec{r}')dV'}{|\vec{r}-\vec{r}'|}$

Then try to get to the following "proof":

$\vec{F}\propto\int_V\frac{\rho(\vec{r}')\nabla\rho(\vec{r})dVdV'}{|\vec{r}-\vec{r}'|}=(...)-\int_V\frac{\rho(\vec{r})\nabla\rho(\vec{r}')dVdV'}{|\vec{r}-\vec{r}'|}\qquad (1)$

The bracket $(...)$ is the integral:

$\int_V\rho(\vec{r})\rho(\vec{r}')dVdV'\nabla\frac{1}{|\vec{r}-\vec{r}'|}=-\int_V\rho(\vec{r})\rho(\vec{r}')dVdV'\frac{\vec{r}-\vec{r}'}{2|\vec{r}-\vec{r}'|^3}=0\qquad (2)$

It is zero because the integrand is an anti-simmetric function with respect to the variable change $\vec{r}\to\vec{r}',\; \vec{r'}\to\vec{r}$. Bythe way, it is your the very first expression for he self-force, but with explicit electric field.

The last integral in (1) is the same thing as my first expression for the force, but with the opposite sign if one makes the variable change $\vec{r}\to\vec{r}',\; \vec{r'}\to\vec{r}$, so it is nothing but zero.

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Here is my thought:

$F_{tot} = \int_{V} \rho\vec{E}dx^3 = \int_{whole space} \rho\vec{E}dx^3$ since charge distribution is bounded in finite region

$\int_{whole space} \rho\vec{E}dx^3 = \int_{whole space}\frac{1}{4\pi}(\vec{\nabla}\vec{E})\vec{E}dx^3$

$\int_{whole space}\frac{1}{4\pi}(\vec{\nabla}\vec{E})\vec{E}dx^3 = \int_{whole space}\frac{1}{8\pi}\vec{\nabla}\vec{E}^2dx^3 = \int_{surface at infinity}\frac{1}{8\pi}\vec{E}^2d\vec{S}$

The last identity comes from another form of Divergence Theorem:

$\int_{V}\vec{\nabla}\psi dx^3 = \int_{S}\psi d\vec{S}$

Now at infinity $\vec{E} \rightarrow 0$, the integral will be $0$. Therefore, self-force vanishes

By the way, it would be funny if you are also in PHY322 :)

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  • $\begingroup$ Haha, hey there! I also first thought of that but the third equality though deceptive, is not true. $\vec{\nabla}\cdot\vec{E}^2=2(\vec{E}\cdot\vec{\nabla})\vec{E}$ $\endgroup$ – Kaan Baykara Jan 21 at 4:08
  • $\begingroup$ I doubt that, in fact you can see the identity I give should hold from Wikipedia link $\endgroup$ – Jiashen Tang Jan 21 at 15:56
  • $\begingroup$ Writing more explicitly, $$((\nabla \cdot \vec{E})\vec{E})= \sum_{i=1}^{3}(\sum_{j=1}^{3} \partial_j E_j) E_i \pmb{e_i}=\sum_{i,j} E_i \partial_j E_j \pmb{e_i}$$ where $\pmb{e_i}$ is the basis. $$(\nabla (E^2))= \sum_{i=1}^{3} \partial_i (\sum_{j=1}^{3} E_j ^2) \pmb{e_i}=\sum_{i,j} E_j \partial_i E_j \pmb{e_i}$$ For example looking at the coefficient of $\pmb{e_1}$ $$E_1 (\sum_{j=1}^{3} \partial_j E_j) \ne \sum_{j=1}^{3} E_j \partial_1 E_j$$ The $\nabla$ in the wikipedia page is not divergence, it is the tensor derivative $\endgroup$ – Kaan Baykara Jan 21 at 22:51

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