Answer
$D = 3.08 \times 10^{-7} C/m^{2}$
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:
dielectric displacement for part a (V = 39.73 V)
Solution:
Using a combination of Equations 18.31, 18.32 and 18.6:
$D = ε_{0}E + P = ε_{0}E + ε_{0}(ε_{r} - 1)E = ε_{0}ε_{r}E = \frac{ ε_{0}ε_{r}V}{l} = \frac{(8.85 \times 10^{-12} F/m)(3.5)(39.73 V)}{(4 \times 10^{-3} m)} = 3.08 \times 10^{-7} C/m^{2}$
