(a)
Given data
*The given charge is
[tex]q=6.0\text{ pc = 6.0}\times10^{-12}\text{ C}[/tex]*The given potential difference is V = 1.25 V
The formula for the capacitance is given as
[tex]C=\frac{q}{V}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} C=\frac{6.0\times10^{-12}}{1.25} \\ =4.8\times10^{-12}\text{ V} \end{gathered}[/tex]Hence, the capacitance is
[tex]C=4.8\times10^{-12}\text{ V}[/tex](b)
Given data
*The given battery voltage is V = 1.50 V
*The given capacitance is
[tex]C=200\text{ nF = 200}\times10^{-9}\text{ F}[/tex]The expression for the electrical potential energy is given as
[tex]U_p=\frac{1}{2}CV^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} U_p=\frac{1}{2}(200\times10^{-9})(1.50)^2 \\ =2.25\times10^{-7}\text{ J} \end{gathered}[/tex]Hence, the electrical potential energy is
[tex]U_p=2.25\times10^{-7}\text{ J}[/tex]