How much energy, in picoJoules, is stored in a parallel plate capacitor connected to a 25.045 V battery? The plates at 18.655 cm x 18.655 cm, and separated by 7.686 mm with a thin sheet of polyethylene between the plates.

Answer :

The energy can be found as follows:

[tex]U=\frac{1}{2}CV^2[/tex]

where:

C = Capacitance

V = Voltage

We can find the Capacitance using the following formula:

[tex]C=\frac{\epsilon_pA}{d}[/tex]

Where:

[tex]\begin{gathered} \epsilon_p=Electric_{\text{ }}permittivity_{\text{ }}of_{\text{ }}polyethylene=2.25 \\ d=distance=7.686\times10^{-3}m \\ A=Area=0.0348m^2 \end{gathered}[/tex]

So:

[tex]C=10.1876F[/tex]

Therefore, the energy is:

[tex]\begin{gathered} E=\frac{1}{2}(10.1876)(25.045^2) \\ E\approx3195.1J \end{gathered}[/tex]