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:

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.