Derive the following equation Px = ε0χ0Ex + 6ε0χ1111E̅2Ex For the polarization induced in an isotropic medium by a linearly polarized electromagnetic wave.
Added by Ethan J.
Step 1
To derive the equation \( P_x = \varepsilon_0 \chi_0 E_x + 6 \varepsilon_0 \chi_{1111} \bar{E}^2 E_x \) for the polarization induced in an isotropic medium by a linearly polarized electromagnetic wave, we will follow these steps: Show more…
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Write a relation for polarisation p of a dielectric material at presence of an external electric field
Adi S.
$$ \begin{aligned} &E_{p}=\frac{q \overrightarrow{r_{1}}}{4 \pi \varepsilon_{0} r_{1}^{3}}+\frac{q^{\prime} \overrightarrow{r_{2}}}{4 \pi r_{2}^{3} \varepsilon_{0}} ; P \text { in } 1\\ &E_{p}=\frac{q^{\prime \prime} \overrightarrow{r_{1}}}{4 \pi \varepsilon_{0} r_{1}^{3}}, P \text { in 2 }\\ &=6\\ &\text { where } q^{\prime \prime}=\frac{2 q}{\varepsilon+1}, q^{\prime}=q^{\prime \prime}-q\\ &\text { In the limit } \vec{l} \rightarrow 0\\ &\overrightarrow{E_{p}}=\frac{\left(q+q^{\prime}\right) \vec{r}}{4 \pi \varepsilon_{0} r^{3}}=\frac{q \vec{r}}{2 \pi \varepsilon_{0}(1+\varepsilon) r^{3}} \text { , in either part. }\\ &\text { Thus, } \quad E_{p}=\frac{q}{2 \pi \varepsilon_{0}(1+\varepsilon) r^{2}}\\ &\varphi=\frac{q}{2 \pi \varepsilon_{0}(1+\varepsilon) r}\\ &D=\frac{q}{2 \pi \varepsilon_{0}(1+\varepsilon) r^{2}} \times\left\{\begin{array}{c} 1 \text { in vacuum } \\ \varepsilon \text { in dielectric } \end{array}\right. \end{aligned} $$
Electrodynamics
Conductors and Dielectrics in an Electric Field
Let the field in the dielectric be $\vec{E}$ making an angle $\alpha$ with $\vec{n} .$ Then we have the boundary conditions, $E_{0} \cos \alpha_{0}=\varepsilon E \cos \alpha$ and $E_{0} \sin c_{0}=E \sin \alpha$ So $E=E_{0} \sqrt{\sin ^{2} \alpha_{0}+\frac{1}{\varepsilon^{2}} \cos ^{2} \alpha_{0}}$ and $\tan \alpha=\varepsilon \tan \alpha_{0}$ In the dielectric the normal component of the induction vector is $D_{n}=\varepsilon_{0} \varepsilon E_{n}=\varepsilon_{0} \varepsilon E \cos \alpha=\varepsilon_{0} E_{0} \cos \alpha_{0}$ $\sigma^{\prime}=P_{n}=D_{n}-\varepsilon_{0} E_{n}=\left(1-\frac{1}{\varepsilon}\right) \varepsilon_{0} E_{0} \cos \alpha_{0}$ or, $\quad \sigma^{\prime}=\frac{\varepsilon-1}{\varepsilon} \varepsilon_{0} E_{0} \cos \alpha_{0}$
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