Question

The electric potential difference of a 150 µF capacitor is measured across the terminals of the capacitor and found to be 5 volts. What is the potential energy of the capacitor?

Answer 1

it's 2 on edge

Step-by-step explanation:

Answer 2

The potential energy stored in a capacitor is given by:

`U= (1)/(2)CV^2`
where
C is the capacitance of the capacitor
V is the potential difference across the plates of the capacitor

In our problem,

`C=150 \mu F= 150 \cdot 10^(-6) F`

`V=5 V`
Therefore, the potential energy of the capacitor is

`U= (1)/(2)(150 \cdot 10^(-6)F)(5 V)^2 = 1.88 \cdot 10^(-3)J`