- #1

Kaan99

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## Homework Statement

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

## Homework Equations

[itex]-\vec{\nabla}\Phi=\vec{E}[/itex]

[itex]\vec{\nabla}\cdot \vec{E}=\frac{\rho}{\epsilon_0}[/itex]

## 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 [itex]\rho=0[/itex] 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.