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The inner radius of the dielectric shell is $a$, and the outer radius is $b$. I tried to use the formula involving the square of the electric field, and found $E$ in three regions. But when I evaluate the integral due to the first region (that is r from 0 to a), there comes a term 1/0 for the lower limit. I know it's because of the point charge's self-energy, but how can I avoid it? The electrostatic energy of the system just can't be infinite.

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2 Answers 2

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You should subtract the self-energy of the point charge: $U=1/2 \int \epsilon E^2 dv-1/2 \epsilon_0\int E_{\text{point charge}}^2 dv$

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Because of the spherical symmetry, we can apply Gauss Theorem to find $\vec{E}$ and $\vec{D}$ in the three regions.

For $r>b$,

$$\vec{E}=\frac{q}{4\pi\varepsilon_0 r^2}\vec{u}_r$$

For $a<r<b$,

$$\vec{E}=\frac{q}{4\pi\varepsilon r^2}\vec{u}_r$$

For $r<a$,

$$\vec{E}=\frac{q}{4\pi\varepsilon_0 r^2}\vec{u}_r$$

Notice the charge that are created in both faces of the shell.

See this link for make sure that the electrostatic energy of the system is finite

http://www.physicspages.com/2012/11/10/energy-of-conducting-sphere-in-a-dielectric-shell/

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