A 6.00 µF parallel plate capacitor has a charge of +40 μC. What is the potential energy stored in this capacitor?Answer Choices103 μJ113 μJ123 μJ 133 μJ

Respuesta :

In order to calculate the energy stored in a capacitor, we can use the formula below:

[tex]E=\frac{Q^2}{2C}[/tex]

Where Q is the charge and C is the capacitance.

So, for Q = 40 * 10^-6 C and C = 6 * 10^-6 F, we have:

[tex]\begin{gathered} E=\frac{(40\cdot10^{-6})^2}{2\cdot6\cdot10^{-6}} \\ E=\frac{1600\cdot10^{-12}}{12\cdot10^{-6}} \\ E=133\cdot10^{-6}\text{ J} \\ E=133\text{ }\mu J \end{gathered}[/tex]

Therefore the correct option is the fourth one.

Otras preguntas