Question

25.4 energy stored in an electric field a circuit consists of a battery with an emf of 1.50 v and three capacitors in parallel, where c1 = 3.00 μf, c2 = 5.00 μf, and c3 = 8.00 μf. what energy is stored in the capacitors?

Explanation:

Step1: Calculate equivalent capacitance

For capacitors in parallel, the equivalent capacitance $C_{eq}=C_1 + C_2+C_3$. Given $C_1 = 3.00\ \mu F$, $C_2=5.00\ \mu F$, $C_3 = 8.00\ \mu F$, so $C_{eq}=3.00 + 5.00+8.00=16.00\ \mu F=16\times10^{- 6}\ F$.

Step2: Use energy - storage formula

The energy stored in a capacitor is given by $U=\frac{1}{2}CV^{2}$. Here, $C = C_{eq}$ and $V = 1.50\ V$. Substitute the values: $U=\frac{1}{2}\times(16\times10^{-6})\times(1.50)^{2}$.
$U=\frac{1}{2}\times16\times10^{-6}\times2.25 = 18\times10^{-6}\ J = 18\ \mu J$.

Answer:

$18\ \mu J$