Question

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

Answer 1

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

`E=(Q^2)/(2C)`

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:

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

Therefore the correct option is the fourth one.