A parallel plate capacitor has a plate spacing of 1.5 mm, which is filled with a dielectric of κ = 4.3, and its capacitance is 80 μF. If the dielectric is taken out and the plate spacing doubled, then what is the new capacitance?

Question

Quizplus · Accepted Answer

The original capacitance with the dielectric is given by C = κε₀A/d. Removing the dielectric and doubling the plate spacing gives the new capacitance as C' = ε₀A/(2d). The ratio of the new capacitance to the original is C'/C = 1/(2κ). Solving for C' gives C' = 80 μF / (2 * 4.3) = 9.3 μF.

A Parallel Plate Capacitor Has a Plate Spacing of 1