Problem 3.78

Near the point $A$ (Fig. 3.10) lying on the boundary between glass and vacuum the electric field strength in vacuum is equal to $E_{0}=10.0 \mathrm{~V} / \mathrm{m},$ the angle between the vector $\mathbf{E}_ {0}$ and the normal n of the boundary line being equal to $\alpha_{0}=30^{\circ} .$ Find the field strength $E$ in glass near the point $A,$ the angle $\alpha$ between the vector $\mathbf{E}$ and $\mathbf{n},$ as well as the surface density of the bound charges at the point $A$ .

\begin{aligned} &\text { 3.78. } E=\frac{E_{0}}{\varepsilon} \sqrt{\cos ^{2} \alpha_{0}+\varepsilon^{2} \sin ^{2} \alpha_{0}}=5.2 \mathrm{~V} / \mathrm{m}\\ &\tan \alpha=\varepsilon \tan \alpha_{0}, \text { hence, } \alpha=74^{\circ} ; \sigma^{\prime}=\frac{\varepsilon_{0}(\varepsilon-1)}{\varepsilon} E_{0} \cos \alpha_{0}=64 \mathrm{pC} / \mathrm{m}^{2} \end{aligned}

Conductors and Dielectrics in an Electric Field