Materials Science and Engineering: An Introduction

$l_{2} = 3.36 mm$
Given: A parallel-plate capacitor using a dielectric material having an $ε_{r}$ of 2.2 has a plate spacing of 2 mm (0.08 in.). Required: The new spacing between the plates if another material having a dielectric constant of 3.7 is used and the capacitance is to be unchanged Solution: Using a combination of Equations 18.26 and 18.27, we find: $C = \frac{εA}{l} = \frac{ε_{r}ε_{0}A}{l}$ Since $C_{1} = C_{2}$, let subscripts 1 and 2 be the initial and final state: $\frac{ε_{r1}ε_{0}A}{l_{1}} = \frac{ε_{r2}ε_{0}A}{l_{2}}$ $l_{2} = \frac{ε_{r2}l_{1}}{ε_{r1}} = \frac{(3.7)(2 mm)}{2.2} = 3.36 mm$