If a dielectric with relative permittivity ε_r fills the space between the plates of a parallel-plate capacitor, the capacitance changes by what factor?

If a dielectric with relative permittivity ε_r fills the space between the plates of a parallel-plate capacitor, the capacitance changes by what factor?

• A. It remains equal to the vacuum capacitance
• B. It increases by a factor ε_r
• C. It decreases by a factor ε_r
• D. It increases by a factor ε_r^2

Answer: (B)

The capacitance of a parallel-plate capacitor scales with the material's permittivity between the plates. For a plate area A and separation d, C = εA/d, where ε = ε0εr. Without a dielectric, ε = ε0 and the capacitance is C0 = ε0A/d. With the dielectric, C = ε0εrA/d = εrC0. So the capacitance increases by a factor εr. The dielectric’s polarization makes it easier to store charge at the same voltage, or equivalently, for the same charge the voltage drops by a factor εr.

The capacitance of a parallel-plate capacitor scales with the material's permittivity between the plates. For a plate area A and separation d, C = εA/d, where ε = ε0εr. Without a dielectric, ε = ε0 and the capacitance is C0 = ε0A/d. With the dielectric, C = ε0εrA/d = εrC0. So the capacitance increases by a factor εr. The dielectric’s polarization makes it easier to store charge at the same voltage, or equivalently, for the same charge the voltage drops by a factor εr.