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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 4.14

Let's start by recalling that the bound charges are given by (Given some polarization $\mathbf{P}$)

$${\sigma }_b=\mathbf{P}\cdot \hat{\mathbf{n}}$$

And

$${\rho }_b=-\mathrm{\nabla }\cdot \mathbf{P}$$

Where the first denotes surface (bound) charge (density) and the latter volume (bound) charge (density).

Now that we have the (bound) densities, the total bound charge is given by integrals of the aforementioned densities:

$$Q_{\text{total,bound}}={\oint }_{\mathcal{S}}{\sigma }_{b}da+{\int }_{\mathcal{V}}{\rho }_{b}d\tau$$

But we know what ${\sigma }_b$ and ${\rho }_b$ are, so let's replace them in the above to get:

$$Q_{\text{total,bound}}={\oint }_{\mathcal{S}}\mathbf{P}\cdot \hat{\mathbf{n}}-{\int }_{\mathcal{V}}\mathrm{\nabla }\cdot \mathbf{P}d\tau$$

But the divergence theorem tells us that closed surface integrals are related to volume integrals through a divergence:

$${\int }_{\mathcal{V}}\mathrm{\nabla }\cdot \mathbf{P}d\tau ={\oint }_{\mathcal{S}}\mathbf{P}\cdot d\mathbf{a}$$

Therefore, the above expression vanishes and we conclude that

$$Q_{\text{total,bound}}=0$$

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