Edits to “Statement”, “Solution”, “Answer”

jzmicer edited
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revision #19256 parent #19255 ← older
en/6.6.12.md +68 −4
@@ -1,11 +1,75 @@
### Statement
−$6.6.12.$ [Insert the problem statement]
+$6.6.12.$ The space between the plates of a parallel‑plate capacitor is filled with two layers of different dielectrics of thicknesses $d_1$ and $d_2$. The dielectric permittivities of the dielectrics are $\varepsilon_1$ and $\varepsilon_2$. The plate area is $S$. Find the capacitance of the capacitor. What charge will be induced at the interface between the dielectrics if a charge $\pm q$ is placed on the capacitor plates?
+
+![|351x158, 50%](../../img/6.6.12/6.6.12.png)
+
### Solution
−1
+Our capacitor can be represented as two capacitors connected in series; for them the capacitance is
+$$
+\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\tag{1}
+$$
+where
+$$
+C_i=\frac{\varepsilon_i\varepsilon_0S}{d_i}\tag{2}
+$$
+$(2)\to(1):$
+$$
+\frac{1}{C}=\frac{1}{\varepsilon_0S}\left(\frac{d_1}{\varepsilon_1}+\frac{d_2}{\varepsilon_2}\right)
+$$
+$$
+C=\frac{\varepsilon_0\varepsilon_1\varepsilon_2S}{\varepsilon_2d_1+\varepsilon_1d_2}\tag{3}
+$$
−#### Answer
+Now recall that at the interface between dielectrics the normal component of the field undergoes a jump. The dielectrics are isotropic, the field is normal to the interface, and there are no free charges at the interface, so
+$$
+\varepsilon_2E_2=\varepsilon_1E_1\tag{4}
+$$
+For the voltage across the capacitor after placing the charge:
+$$
+V=E_1d_1+E_2d_2=\frac{q}{C}\tag{5}
+$$
+The surface density of bound (induced) charges at the interface is determined by the jump in the normal component of the polarisation vector:
+$$
+\sigma'=P_{n1}-P_{n2}\tag{6}
+$$
+For an isotropic dielectric
+$$
+\vec P=(\varepsilon-1)\varepsilon_0\vec E
+$$
+Therefore
+$$
+\sigma'=(\varepsilon_1-1)\varepsilon_0E_1-(\varepsilon_2-1)\varepsilon_0E_2
+=\varepsilon_0[(\varepsilon_1E_1-\varepsilon_2E_2)+E_2-E_1]
+$$
+Taking $(4)$ into account, we finally obtain a nice expression:
+$$
+\sigma'=\varepsilon_0(E_2-E_1)\tag{7}
+$$
−[Insert a concise answer or boxed result]
+Solve the system $(4), (5), (7)$:
+$$
+E_1=\frac{q\varepsilon_2}{C(d_1\varepsilon_2+d_2\varepsilon_1)}, \quad
+E_2=\frac{\varepsilon_1}{\varepsilon_2}E_1=\frac{q\varepsilon_1}{C(d_1\varepsilon_2+d_2\varepsilon_1)}
+$$
+$$
+\sigma'=\varepsilon_0q\frac{\varepsilon_1-\varepsilon_2}{C(d_1\varepsilon_2+d_2\varepsilon_1)}
+=\varepsilon_0q\frac{\varepsilon_2d_1+\varepsilon_1d_2}{\varepsilon_0\varepsilon_1\varepsilon_2S}\,
+\frac{\varepsilon_1-\varepsilon_2}{d_1\varepsilon_2+d_2\varepsilon_1}
+$$
+$$
+\sigma'=\frac{q'}{S}=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}\frac{q}{S}
+$$
+$$
+q'=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}q
+$$
+
+#### Answer
+$$
+C=\frac{\varepsilon_0\varepsilon_1\varepsilon_2S}{\varepsilon_2d_1+\varepsilon_1d_2}
+$$
+$$
+q'=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}q
+$$