Answer:
(a) Approximately [tex]10.1\; {\rm V}[/tex].
Explanation:
Let [tex]C[/tex] denote the capacitance of a capacitor. Let [tex]V[/tex] be the potential difference (voltage) between the two plates of this capacitor. The energy [tex]E[/tex] stored in this capacitor would be:
[tex]\displaystyle E = \frac{1}{2}\, C\, (V^{2})[/tex].
Rearrange this equation to find an expression for the potential difference [tex]V[/tex] in terms of capacitance [tex]C[/tex] and energy [tex]E[/tex]:
[tex]\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}[/tex].
[tex]\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}[/tex]
The capacitance [tex]C[/tex] of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):
[tex]\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}[/tex].
Given that the energy stored in this capacitor is [tex]E = 1.85 \times 10^{-5}\; {\rm J}[/tex], the potential difference across the capacitor plates would be:
[tex]\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}[/tex].
