Final answer:
Pulling the plates of an isolated charged capacitor apart decreases the capacitance and increases the potential difference. Increasing the area of the plates while keeping the voltage constant increases both the capacitance and the stored energy. The addition of a dielectric material can also increase capacitance by reducing the electric field strength.The correct option is B.
Step-by-step explanation:
Understanding Capacitance and Potential in Capacitors
When the plates of an isolated charged capacitor are pulled apart, it has a specific effect on the capacitance and potential difference of the capacitor. The capacitance (C) of a parallel plate capacitor is given by the equation C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. As the distance between the plates increases, the capacitance decreases since capacitance is inversely proportional to the distance.
Given that capacitance is defined as C = Q/V, with Q being the charge and V the potential difference, and since the charge on an isolated capacitor remains constant, as the capacitance decreases the potential difference must increase to maintain the relationship defined by the equation. Therefore, pulling the plates of an isolated charged capacitor apart decreases the capacitance and increases the potential difference.
Furthermore, if the capacitor in question were to increase its plate area while maintaining a constant voltage, the capacitance would increase, because a larger area A results in a higher capacitance. As the capacitance increases, for a fixed voltage, the amount of charge (Q) it can store also increases, which means the stored energy in the capacitor, given by ½ CV2, would also increase. Therefore, increasing the area while the voltage is held fixed results in an increase in both capacitance and stored energy.
Additionally, when discussing the addition of a dielectric material, the insertion of a dielectric between the plates of a capacitor reduces the electric field strength (E) and therefore reduces the potential difference (V = Ed) across the plates, which results in an increase in capacitance since the charge remains the same (C = Q/V).
The correct option is B.