Physics
Physics
Problem
2. A parallel-plate capacitor is charged by connection to a battery. If the battery is disconnected and the separation between the plates is increased, what will happen to the charge on the capacitor and the voltage across it? (A) Both remain fixed. (B) Both increase. (C) The charge increases and the voltage decreases. (D) The charge remains fixed and the voltage increases. (A) Both remain fixed. (B) Both increase. (C) The charge increases and the voltage decreases. (D) The charge remains fixed and the voltage increases.Intuitive explanation
When you disconnect the battery, the charge on the plates becomes trapped—there's nowhere for it to go. The key insight is that capacitance depends on plate separation, so increasing decreases . Since and is constant while shrinks, the voltage must increase. Think of it like squeezing the same amount of charge into a "weaker" capacitor.
Intuitive explanation
When you disconnect the battery, the charge on the plates becomes trapped—there's nowhere for it to go. The key insight is that capacitance depends on plate separation, so increasing decreases . Since and is constant while shrinks, the voltage must increase. Think of it like squeezing the same amount of charge into a "weaker" capacitor.