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: $$\frac{1}{C_{total}} = \frac{h/4}{\varepsilon_0 S} + \frac{h/4}{\varepsilon_0 S}.$$
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: $B = -\frac{\mu_0 \varepsilon_0 v V}{h}$.
$b.$ It decreases by a factor of $\frac{\varepsilon + 1}{\varepsilon - 1}$.