A parallel plate capacitor has a plate spacing of 1.5 mm,which is filled with a dielectric of $\kappa$ = 4.3,and its capacitance is 80 $\mu$F.If the dielectric is taken out and the plate spacing doubled,then what is the new capacitance?
A)690 $\mu$F
B)37 $\mu$F
C)9.3 $\mu$F
D)170 $\mu$F
E)19 $\mu$F

Question

Quizplus · Accepted Answer

The original capacitance with the dielectric is given by C = ε₀κA/d. When the dielectric is removed and the spacing is doubled, the new capacitance becomes C' = ε₀A/(2d). The ratio of the new capacitance to the original is 1/(2κ), leading to C' = 80 μF / (2 * 4.3) = 9.3 μF.

A Parallel Plate Capacitor Has a Plate Spacing of 1 κ\kappaκ

A Parallel Plate Capacitor Has a Plate Spacing of 1 $\kappa$