If a dielectric with dielectric constant κ completely fills the space between the plates of a parallel-plate capacitor, by what factor does the capacitance increase compared to vacuum?

If a dielectric with dielectric constant κ completely fills the space between the plates of a parallel-plate capacitor, by what factor does the capacitance increase compared to vacuum?

• A. κ
• B. 1/κ
• C. κ^2
• D. Remains unchanged

Answer: (A)

Putting a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant. Capacitance is C = Q/V. For a parallel-plate capacitor in vacuum, C0 = ε0 A/d. When a dielectric with dielectric constant κ fills the space, the electric displacement is D = ε0 κ E, so the charge on the plates is Q = ∮ D·dA = ε0 κ E A. With E = V/d, this gives Q = ε0 κ (V/d) A, and thus C = Q/V = ε0 κ A/d = κ C0. Therefore the capacitance becomes κ times larger than in vacuum. This rules out 1/κ, κ^2, or no change.

Putting a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant. Capacitance is C = Q/V. For a parallel-plate capacitor in vacuum, C0 = ε0 A/d. When a dielectric with dielectric constant κ fills the space, the electric displacement is D = ε0 κ E, so the charge on the plates is Q = ∮ D·dA = ε0 κ E A. With E = V/d, this gives Q = ε0 κ (V/d) A, and thus C = Q/V = ε0 κ A/d = κ C0. Therefore the capacitance becomes κ times larger than in vacuum. This rules out 1/κ, κ^2, or no change.