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@@ -1,11 +1,62 @@
### Statement
−$11.6.9.$ [Insert the problem statement]
+$11.6.9.$ Inside a parallel‑plate capacitor, parallel to its plates, a conducting plate of thickness equal to half the distance between the capacitor plates moves with velocity $v$. The voltage across the capacitor plates is maintained at $V$, and the separation between them is $h$.
+$a.$ What is the magnetic induction inside the conductor? Between the moving conductor and the capacitor plates?
+
+$b.$ How does the magnetic induction inside the plate change if the conductor is replaced by a dielectric with dielectric permittivity $\varepsilon$?
+
### Solution
+$a.$ The system can be represented as two capacitors:
+$$
+\frac{1}{C_{total}} = \frac{h/4}{\varepsilon_0 S} + \frac{h/4}{\varepsilon_0 S}.
+$$
−1
+The field inside the capacitor:
+$$
+E = \frac{\sigma}{\varepsilon_0} = \frac{q}{\varepsilon_0 S} = \frac{C_{total} V}{\varepsilon_0 S} = \frac{2V}{h}.
+$$
+The conducting plate completely "expels" the field inside itself, creating an oppositely directed field of the same magnitude. Consider the region that the plate has shifted into during time $dt$ (which has become inside the plate during $dt$):
+$$
+\oint \vec B \, d\vec l = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \int \vec E \, d\vec S \right),
+$$
+$$
+B \cdot 2l = \mu_0 \varepsilon_0 v l E,
+$$
+$$
+B = \frac{\mu_0 \varepsilon_0 v V}{h}.
+$$
+
+Now consider the space between the plate and the conductor. Before the plate appears and after, the total capacitance can be represented as:
+$$
+C_0 = \frac{\varepsilon_0 S}{h}, \qquad C_{total} = \frac{2\varepsilon_0 S}{h}.
+$$
+Then displacement currents arise in the space to change the voltage in accordance with the new capacitance. The field has increased by a factor of 2, which means a current appears whose magnitude is equal to that obtained earlier and whose direction is opposite (i.e., the induction has a minus sign).
+
+$b.$
+The field inside the dielectric does not vanish, but decreases by a factor of $\varepsilon$. Taking into account that the voltage on the capacitor plates is constant, the field inside changes compared to part $a$ due to the change in capacitance:
+$$
+\frac{1}{C_{total}} = \frac{h/4}{\varepsilon_0 S} + \frac{h/4}{\varepsilon_0 S} + \frac{h/2}{\varepsilon_0 \varepsilon S} = \frac{h(\varepsilon + 1)}{2\varepsilon_0 \varepsilon S},
+$$
+$$
+E = \frac{q}{\varepsilon_0 S} = \frac{C_{total} V}{\varepsilon_0 S} = \frac{2V\varepsilon}{h(\varepsilon + 1)}.
+$$
+$$
+\oint \vec B \, d\vec l = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \int \vec E \, d\vec S \right),
+$$
+$$
+B \cdot 2l = \mu_0 \varepsilon_0 v l \left( E - \frac{E}{\varepsilon} \right) = \mu_0 \varepsilon_0 v l E \left( \frac{\varepsilon - 1}{\varepsilon} \right),
+$$
+$$
+B = \frac{\mu_0 \varepsilon_0 v V}{h} \left( \frac{\varepsilon - 1}{\varepsilon + 1} \right) = B_a \left( \frac{\varepsilon - 1}{\varepsilon + 1} \right).
+$$
+
#### Answer
+$a.$ Inside the conductor:
+$B = \frac{\mu_0 \varepsilon_0 v V}{h}$,
+between the conductor and the capacitor plates:
+$B = -\frac{\mu_0 \varepsilon_0 v V}{h}$.
−[Insert a concise answer or boxed result]
+
+$b.$ It decreases by a factor of $\frac{\varepsilon + 1}{\varepsilon - 1}$.