# Electrostatic self-force is zero

1. Homework Statement
Consider a charged body of finite size, ($\rho=0$ outside a bounded region $V$). $\vec{E}$ is the electric field produced by the body. Suppose $\vec{E} \rightarrow 0$ at infinity. Show that the total self-force is zero: $\int_V \rho \vec{E} dV = \vec{0}$, i.e. the charged body does not exert a net force on itself.

2. Homework Equations
$-\vec{\nabla}\Phi=\vec{E}$
$\vec{\nabla}\cdot \vec{E}=\frac{\rho}{\epsilon_0}$

3. The Attempt at a Solution
Integration by parts
$$\int_V \rho \vec{E}\, dV = \int_V -\vec{\nabla}(\rho\Phi)+\Phi \vec{\nabla}\rho \,dV$$
By stokes theorem
$$\int_V \vec{\nabla}(\rho\Phi)=\int_{\partial V}\rho\Phi \hat{n} \,dS$$
If we choose the integration region outside of the charged body (it's possible because the body has finite size), then $\rho=0$ on the boundary of the integration region. Therefore,
$$\int_{\partial V}\rho\Phi \hat{n} \,dS=0$$
$$\int_V \rho \vec{E} \,dV = \int_V\Phi \vec{\nabla}\rho \,dV$$
Beyond this point I've tried using other vector calculus identities, but they were of no use.

You could use $\vec{\nabla}^2 \cdot \Phi=\frac{\rho}{\epsilon_0}$ to replace $\rho$ but I'm not sure if that leads to something useful.
I would have tried to use the symmetry you have. The electric field is produced by the charge. Replace $\vec E$ by an integral over the volume, and then show that the integral is equal to the negative of itself.