Answer
(a) $5.0\times 10^{-6}m$
(b) $67V$
Work Step by Step
(a) We can find the required height as follows:
$\frac{1}{2}CV^2=mgh$
This can be rearranged as:
$h=\frac{CV^2}{2mg}$
We plug in the known values to obtain:
$h=\frac{(0.22\times 10^{-6}F)(1.5V)^2}{2(5\times 10^3Kg)(9.8m/s^2)}$
$h=5.0\times 10^{-6}m$
(b) We can find the required voltage as follows:
$\frac{1}{2}CV^2=mgh$
This can be rearranged as:
$V=\sqrt{\frac{2mgh}{C}}$
We plug in the known values to obtain:
$V=\sqrt{\frac{2(5\times 10^{-3}Kg)(9.8m/s^2)(0.01m)}{0.22\times 10^{-6}F}}$
$V=67V$
