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

Answer:

it's 2 on edge

Explanation:

Answer 2

The potential energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
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
$U= \frac{1}{2}(150 \cdot 10^{-6}F)(5 V)^2 = 1.88 \cdot 10^{-3}J$