Answer
(a) $5.8\times 10^{-14}J$
(b) decrease
Work Step by Step
(a) We can find the energy stored as follows:
$U=\frac{1}{2}(\frac{K\epsilon_{\circ}A}{d})V^2$
We plug in the known values to obtain:
$U=\frac{1}{2}(\frac{(4.5)(8.85\times 10^{-12}C^2/Nm^2)(4.75\times 10^{-9}m^2)}{8.5\times 10^{-9}m})(0.0725V)^2$
$U=5.8\times 10^{-14}J$
(b) We know that $U=\frac{1}{2}(\frac{K\epsilon_{\circ}A}{d})V^2$. This equation shows that the energy and thickness of the membrane are inversely proportional. Thus, if the thickness is increased then the energy will decrease.
