Answer
$V = 39.73 V$
Work Step by Step
Given:
A charge (Q) of $2.0 \times 10^{-10} C $ is to be stored on each plate of a parallel-plate capacitor having an area (A) of $650 mm^{2} (1.0 in^{2}) $ and a plate separation (l) of 4.0 mm (0.16 in.).
Required:
Voltage given a material having a dielectric constant ($ε_{r}$) of 3.5 positioned within the plates.
Solution:
Using the combination of Equations 18.24, 18.26 and 18.27:
$C = \frac{Q}{V} = ε\frac{A}{l} = ε_{r}ε_{0}\frac{A}{l}$
Solving for V:
$V = \frac{Ql}{ε_{r}ε_{0}A} = \frac{(2.0 \times 10^{-10} C )(4 \times 10^{-3} m)}{(3.5)(8.85 \times 10^{-12} F/m)(650 mm^{2})(1 m^{2}/10^{6}mm^{2})} = 39.73 V$
