A capacitor consists of two 6.0-cm-diameter circular plates separated by 1.0 mm. The plates are...

We are given:
${\displaystyle D=6cm}$, ${\displaystyle d=1mm}$, ${\displaystyle V=150V}$
a) The energy stored in the capacitor will be obtained using this formula:
${\displaystyle U=\frac{1}{2}C{V}^2}$
Where ${\displaystyle C}$ is the capacitance of the capacitor and ${\displaystyle D}$ is the potential
The capacitance we can get from the following formula:
${\displaystyle C=\frac{{ϵ}_{0}A}{d}}$
A capacitor consists of circular plates, so its surface is:
${\displaystyle A=\pi \times r^2}$
And radius is:
${\displaystyle r=\frac{D}{2}=\frac{6}{2}=3cm}$
${\displaystyle {ϵ}_0}$ is the dialectric constznt of vacuum and has the following value:
${\displaystyle {ϵ}_0=8.854\times {10}^{-12}\frac{F}{m}}$
Thus, the first formula becomes:
${\displaystyle U=\frac{1}{2}\times \frac{{ϵ}_0\pi r^2}{d}\times V^2}$
Now, convert cm and mm into m
Finally, we have:
${\displaystyle U=\frac{1}{2}\times \frac{8.854\times {10}^{-12}\pi \times {(3\times {10}^{-2})}^2}{1\times {10}^{-3}}\times {\left(150\right)}^2}$
${\displaystyle U=0.28\times {10}^{-6}J}$
b)Here the distance between the plates increases to ${\displaystyle d_1=2mm}$
The work required to achieve will be obtained as the energy difference:
${\displaystyle W=U-U_1}$
${\displaystyle U_1=\frac{1}{2}\times \frac{8.854\times {10}^{-12}\pi \times {(3\times {10}^{-2})}^2}{2\times {10}^{-3}}\times {150}^2=0.14\times {10}^{-6}J}$
And we have:
${\displaystyle W=0.28\times {10}^{-6}-0.14\times {10}^{-6}}$
${\displaystyle W=0.14\times {10}^{-6}J}$