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Capacitors are devices that store electrical energy in an electric field. When connected in parallel, the overall capacitance of the circuit is the sum of the individual capacitances. This is because parallel capacitors collectively increase the ability of a circuit to store charge.

For two capacitors, with capacitances of \(2C\) and \(C\), the total capacitance in parallel is given by:

• \(C_{\text{total}} = 2C + C = 3C\)

Here, each capacitor will face the same potential difference since the voltage across parallel components is always equal, thereby simplifying the charge calculation.

Understanding how capacitance adds up in parallel circuits can help predict how the total stored charge and resulting electric capacity of the circuit will behave during operation.

Charge conservation is a fundamental principle in physics that states the total electric charge in an isolated system remains constant over time. This principle is particularly crucial when the system undergoes rearrangements, like removing the battery after charging the capacitors.

In the exercise, with the battery removed, the total initial charge before the rearrangement is equal to the total final charge. Initially, charges on two capacitors are:

• \(Q_1 = 2CV\)
• \(Q_2 = CV\)

Thus, the total charge \(Q_{\text{total}}\) is \(3CV\), which must be conserved.

This principle allows us to solve for the new potential difference once a dielectric is introduced, as the total charge remains the same across the system, though redistributed as necessary.

A dielectric is an insulating material that, when placed inside a capacitor, affects its capacitance by reducing the electric field, thereby allowing it to store more charge. This is quantified by the dielectric constant \(K\).

When the capacitor of capacity \(C\) is filled with a dielectric, its capacitance changes to \(KC\). Essentially, the dielectric constant influences how much of an electric field a dielectric medium allows. If the dielectric constant is greater than one, the capacitance increases.

Because the task involves taking out the battery and modifying just one capacitor while the total charge is conserved, understanding the role of dielectrics helps explain why the system's potential difference changes after dielectric introduction.

Potential difference, often referred to as voltage, is the energy difference per charge unit between two points in a circuit. In this problem, even after the battery is removed, the capacitors must equilibrate to the same potential difference.

Given that charge conservation holds, and capacitance affects how this charge distributes, the new potential difference \(V'\) is determined by combining the effects of original capacitance and changes due to the dielectric material. This relationship is expressed as:

• \(V' = \frac{3CV}{KC + 2C}\)

Simplifying, we find that the voltage now is \(\frac{3V}{K + 2}\). This shift reflects how both the total charge and alteration in capacitance from the dielectric influence the electric potential in the system.