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+### Statement
+
+$11.2.8.$ [Insert the problem statement]
+
+### Solution
+
+a) Well, everything is pretty trivial here, the voltage on each capacitor:
+
+$$U = \frac{\varphi}{2}$$
+
+So there is a charge on each capacitor:
+
+$$q_1 = C_1 \frac{\varphi}{2}$$
+
+$$q_2 = C_2 \frac{\varphi}{2}$$
+
+b) Let's assume that the field induction is directed inside the drawing. Then the resulting current is shown in the figure:
+![To task $11.2.8$ |1657x1223, 31%](../../img/11.2.8/Снимок экрана 2026-06-23 163031.png)
+
+The figure also shows the charge signs on the capacitor plates and the direction of the field in them (over /against the current).
+
+Let's write down the sum of the voltage drops for the left and right semicircles:
+
+$$U_1 + U_3 = \frac{\varphi}{2}$$
+
+$$U_3 - U_2 = - \frac{\varphi}{2}$$
+
+From here:
+
+$$U_1 = \frac{\varphi}{2} - U_3$$
+$$U_2 = U_3 + \frac{\varphi}{2}$$
+
+Nodes a and b are neutral, so the algebraic sum of the charges on the plates connected to the nodes is zero.:
+
+$$q_1 - q_2 - q_3 = 0$$
+
+$$C_1 U_1 - C_2 U_2 = C_3 U_3$$
+
+Let's substitute our relations for the stresses:
+
+$$C_1\frac{\varphi}{2} - C_1 U_3 - C_2 U_3 - C_2 \frac{\varphi}{2} = C_3 U_3$$
+
+$$U_3 = \frac{C_1 - C_2}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$
+
+$$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$
+
+Savchenko apparently chose the opposite direction of induction, so the answer turned out to be a minus.
+
+#### Answer
+
+[Insert a concise answer or boxed result]