Правка разделов «Statement», «Solution», «Answer»

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@@ -1,11 +1,30 @@
### Statement
+$11.6.2.$ $a.$
−$11.6.2.$ [Insert the problem statement]
+A parallel‑plate capacitor moves with velocity $v$, as shown in the figure. The electric field strength between the plates is $E$. Determine the rate of change of the electric flux through the rectangular contour $abcd$ and the circulation of the magnetic induction around this contour. How are the sought quantities related to each other in SI? In CGS?
+$b.$ Give examples confirming the proportionality of the circulation of the magnetic induction around a contour to the rate of change of the electric flux through the surface bounded by this contour.
+
### Solution
+![|633x341, 60%](../../img/11.6.2/11.6.2.png)
+$a.$
+$$
+\frac{dN}{dt} = E \frac{dS}{dt} = v l E.
+$$
+Choose a surface passing through the capacitor:
+$$
+C_B = \int \vec B \, d\vec l = \mu_0 \sum I = \mu_0 \frac{\sigma l \cdot v \, dt}{dt} = \mu_0 \varepsilon_0 v l E.
+$$
+Using the solution of problem 11.6.1, this can be generalised for any surface:
+$$
+C_B = \mu_0 \varepsilon_0 \frac{dN}{dt} \quad \text{(in SI)}, \qquad C_B = \frac{1}{c} \frac{dN}{dt} \quad \text{(in CGS)}.
+$$
+$b.$
+Charging capacitor:
−1
+When a capacitor is charging, current flows in the wires, but there is no current between the plates. The changing electric field between the plates maintains the circulation of $B$ around a contour that does not intersect the conductor, equal to the circulation around a contour that does intersect the conductor: $C_B = \mu_0 I$ for a surface crossing the conductor.
#### Answer
−
−[Insert a concise answer or boxed result]
+$$
+\boxed { a. \ \frac{dN}{dt} = v l E, \quad C_B = \mu_0 \varepsilon_0 v l E, \quad C_B = \mu_0 \varepsilon_0 \frac{dN}{dt} \ \text{(in SI)}, \quad C_B = \frac{1}{c} \frac{dN}{dt} \ \text{(in CGS)}. }
+$$