A dielectric with constant κ is inserted into a capacitor that is isolated (no connection to a battery), so the charge Q remains fixed. What happens to the capacitance C and the voltage V across the plates?

A dielectric with constant κ is inserted into a capacitor that is isolated (no connection to a battery), so the charge Q remains fixed. What happens to the capacitance C and the voltage V across the plates?

• A. C increases to κ C0; V decreases to V0/κ
• B. C decreases to C0/κ; V increases to κ V0
• C. C unchanged; V unchanged
• D. C increases to κ^2 C0; V decreases to V0/κ^2

Answer: (A)

When a dielectric fully fills the space between the plates of a capacitor, the capacitance increases by a factor equal to the dielectric constant κ. This is because the dielectric makes it easier to store charge for a given voltage, effectively increasing C from its vacuum value C0 to κ C0. Since the capacitor is isolated, the charge Q on the plates stays the same. The relationship Q = C V means the voltage must adjust inversely with the capacitance: V = Q/C. With the new capacitance C = κ C0, the voltage becomes V = Q/(κ C0) = V0/κ, where V0 is the original voltage with no dielectric. Physically, the electric field between the plates is reduced by the dielectric (E = V/d and E = σ/ε with ε = κ ε0), so the voltage across the plates drops by the same factor κ while the charge remains fixed. Hence the correct outcome is that the capacitance increases to κ C0 and the voltage decreases to V0/κ.

When a dielectric fully fills the space between the plates of a capacitor, the capacitance increases by a factor equal to the dielectric constant κ. This is because the dielectric makes it easier to store charge for a given voltage, effectively increasing C from its vacuum value C0 to κ C0.

Since the capacitor is isolated, the charge Q on the plates stays the same. The relationship Q = C V means the voltage must adjust inversely with the capacitance: V = Q/C. With the new capacitance C = κ C0, the voltage becomes V = Q/(κ C0) = V0/κ, where V0 is the original voltage with no dielectric.

Physically, the electric field between the plates is reduced by the dielectric (E = V/d and E = σ/ε with ε = κ ε0), so the voltage across the plates drops by the same factor κ while the charge remains fixed. Hence the correct outcome is that the capacitance increases to κ C0 and the voltage decreases to V0/κ.