A capacitor charged to Q0 and V0 is isolated from the source and the dielectric is inserted (κ). What happens to the voltage across the plates?

A capacitor charged to Q0 and V0 is isolated from the source and the dielectric is inserted (κ). What happens to the voltage across the plates?

• A. V = Q0 / (C0 κ)
• B. V = V0
• C. V = V0 / κ
• D. V = Q0 / C0

Answer: (A)

With the capacitor isolated, the charge on the plates stays fixed at Q0. Inserting a dielectric increases the capacitance by a factor κ, so the new capacitance is C = κC0. The voltage across the plates is V = Q/C, so V = Q0/(κC0) = (Q0/C0)/κ = V0/κ. Equivalently, V = Q0/(C0κ). The voltage drops by the factor κ because the dielectric raises the capacitance while the stored charge remains the same.

With the capacitor isolated, the charge on the plates stays fixed at Q0. Inserting a dielectric increases the capacitance by a factor κ, so the new capacitance is C = κC0. The voltage across the plates is V = Q/C, so V = Q0/(κC0) = (Q0/C0)/κ = V0/κ. Equivalently, V = Q0/(C0κ). The voltage drops by the factor κ because the dielectric raises the capacitance while the stored charge remains the same.