Inserting a dielectric material with dielectric constant κ between the plates of a capacitor increases its capacitance by what factor (assuming vacuum capacitance C0)?

Inserting a dielectric material with dielectric constant κ between the plates of a capacitor increases its capacitance by what factor (assuming vacuum capacitance C0)?

• A. C = 1/κ C0
• B. C = κ C0
• C. C = κ^2 C0
• D. C = C0 / κ

Answer: (B)

The main idea is that inserting a dielectric between the plates increases the capacitor’s ability to store charge for a given voltage. A capacitor’s capacitance is C = ε A / d, where ε is the permittivity of the material between the plates. With a dielectric constant κ, the permittivity becomes ε = κ ε0, so the capacitance becomes C = κ ε0 A / d. The vacuum (or air) capacitance is C0 = ε0 A / d. Comparing the two, C = κ C0. So the capacitance increases by a factor κ when the space is fully filled with a dielectric. This assumes uniform filling and the same plate area and separation. If only part of the space is filled, the effective increase would be between 1 and κ depending on the geometry.

The main idea is that inserting a dielectric between the plates increases the capacitor’s ability to store charge for a given voltage. A capacitor’s capacitance is C = ε A / d, where ε is the permittivity of the material between the plates. With a dielectric constant κ, the permittivity becomes ε = κ ε0, so the capacitance becomes C = κ ε0 A / d. The vacuum (or air) capacitance is C0 = ε0 A / d. Comparing the two, C = κ C0. So the capacitance increases by a factor κ when the space is fully filled with a dielectric. This assumes uniform filling and the same plate area and separation. If only part of the space is filled, the effective increase would be between 1 and κ depending on the geometry.