#### Answer

$l_{2} = 3.36 mm$

#### Work Step by Step

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:
$\frac{ε_{r1}ε_{0}A}{l_{1}} = \frac{ε_{r2}ε_{0}A}{l_{2}}$
$l_{2} = \frac{ε_{r2}l_{1}}{ε_{r1}} = \frac{(3.7)(2 mm)}{2.2} = 3.36 mm$