Answer
For part a where V = 39.73 V:
$C = 5.03 \times 10^{-12} F$
For part b where V = 139.07 V:
$C = 1.44 \times 10^{-12} F$
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:
Capacitance for part a and b.
Solution:
For part a, where V = 39.73 V:
$C = \frac{Q}{V} = \frac{(2.0 \times 10^{-10} C )}{39.73} = 5.03 \times 10^{-12} F$
For part b, where V = 139.07 V:
$C = \frac{Q}{V} = \frac{(2.0 \times 10^{-10} C )}{139.07} = 1.44 \times 10^{-12} F$
