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You want something like the Poisson-Boltzmann equation, or its linearized form, the Debye-Huckel equation. To illustrate the effect, consider an immobile spherical charge $Q$ at the center of your coordinate system, surrounded by small mobile charge carriers of charge $\pm q$. Gauss's law will give you the potential $\phi(r)$ as a (spherically symmetric) ...

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I'll give you a simple answer. When you consider the capacitor as two parallel plate capacitors connected in parallel, you will see that their potential differences must be same. But not their charge. The charges on the two capacitors will be different. Thus electric field outside of dielectric in lower part of capacitor is not equal to the electric field ...

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Suppose on the other hand the field in the two places were not equal. Consider a loop integral around the red loop in say anticlockwise direction as shown in the figure. Only the vertical edges contribute to the integral.If $E_1 \neq E_2$,it is obvious that the loop integral is non-zero.This violates the conservative nature of the $\vec{E}$ field in ...

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There are two contributions to the electric field in a dielectric: The field generated by the 'free' charges, i.e the ones on the capacitor plates. Call it $E_0$ $E_0$ polarizes the dielectric, which in turn adds to the total electric field. Call that polarization $P$. The total electric field is $$E=E_0-\epsilon_0^{-1}P$$ (The factor of ...

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From skimming a few articles and patents on e-ink driver technology, my impression is that the primary reason is that each microcapsule acts as a capacitor. Once voltage is applied, the particles move to one electrode or the other and remain there because there is no drain path for the charge. The 'gooiness' of the fluid helps, as evidenced by the typical ...

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The presence of the external electric field breaks $SO(3)$ to $SO(2)$. Suppose that $\mathbf{E}$ is oriented along the $z$-axis, then rotations about the $z$-axis (of course, chosen to pass through the center of the sphere) is a symmmetry in the problem. This $SO(2)$ invariance only implies that the potential is independent of $\varphi$, the polar angle. ...

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