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Electrostatic self-force is zero

  • #1
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1. 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.

2. Homework Equations
[itex]-\vec{\nabla}\Phi=\vec{E}[/itex]
[itex]\vec{\nabla}\cdot \vec{E}=\frac{\rho}{\epsilon_0}[/itex]

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 [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.
 

Answers and Replies

  • #2
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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.
 

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