### Problem StatementTwo identical capacitors (A and B) with capacitance \( C_0 \) are hooked up to a battery of voltage \( V_0 \). A dielectric of constant \( \kappa_0 \) is added between the plates of capacitor B while it is connected to the battery, as shown in the figure.**a)** Is the charge on capacitor A (\( Q_A \)) greater than, less than, or equal to the charge on capacitor B (\( Q_B \))? What about the voltages (\( V_A \) and \( V_B \))? Explain your reasoning.The capacitors above are carefully disconnected from the battery (without being discharged). After the battery is disconnected, the dielectric slab is removed from capacitor B. The capacitors remain connected to each other the entire time.**b)** Determine an expression for the final charges on each of the capacitors (\( Q_{A,f} \) and \( Q_{B,f} \)) after the dielectric has been removed in terms of the given constants (\( \kappa_0, C_0, V_0 \)). Show your work and explain your reasoning.### Explanation of the DiagramThe diagram shows a circuit with two capacitors connected in parallel to a battery with a voltage \( V_0 \). Capacitor A has a capacitance \( C_A = C_0 \), and capacitor B has a capacitance \( C_B = \kappa_0 C_0 \) due to the dielectric. The dielectric is represented as a shaded region within capacitor B.

We are given the 2 capacitors. When they are connected in parallel, the charge on each capacitor is…

Answered: Two identical capacitors (A and B) with…

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