If a capacitor is connected to a fixed voltage source and a dielectric is inserted, what happens to the capacitance and charge?

If a capacitor is connected to a fixed voltage source and a dielectric is inserted, what happens to the capacitance and charge?

• A. Capacitance remains C; voltage fixed; charge unchanged
• B. Capacitance increases to C' = κ C; voltage fixed; charge increases to Q' = C' V
• C. Capacitance decreases to C/κ; voltage fixed; charge decreases
• D. Capacitance increases to κ C; voltage increases to κ V

Answer: (B)

Inserting a dielectric between the plates increases the capacitor’s ability to store charge at a given voltage. The capacitance becomes larger by the dielectric constant κ, so C' = κC. Since the voltage is fixed by the source, the charge must adjust to Q' = C'V. With C' = κC, that gives Q' = κCV = κQ. So the capacitance increases by κ, the voltage remains the same, and the charge increases by the same factor κ. The stored energy U = 1/2 C V^2 also increases by κ because V is unchanged while C grows.

Inserting a dielectric between the plates increases the capacitor’s ability to store charge at a given voltage. The capacitance becomes larger by the dielectric constant κ, so C' = κC. Since the voltage is fixed by the source, the charge must adjust to Q' = C'V. With C' = κC, that gives Q' = κCV = κQ. So the capacitance increases by κ, the voltage remains the same, and the charge increases by the same factor κ. The stored energy U = 1/2 C V^2 also increases by κ because V is unchanged while C grows.