3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?

Respuesta :

Answer:

9.72 x 10^(-11)

Explanation:

The energy stored in a capacitor can be calculated as:

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

Where C is the capacitance and V is the Voltage. So, replacing C by 1.35pF or 1.35 x 10^(-12) F, and V by 12.0 V, we get:

[tex]E=\frac{1.35\times10^{-12}F(12.0V)^2}{2}=9.72\times10^{-11}J[/tex]

Therefore, the capacitot would store 9.72 x 10^(-11) J of energy.