Question

a 20 - μf capacitor is charged to 200 v. its stored energy is 4 j 0.4 j 0.1 j 0.004 j 4000 j save for later 25.4 energy stored in an electric field

Explanation:

Step1: Recall energy - formula for capacitor

The energy stored in a capacitor is given by $U=\frac{1}{2}CV^{2}$, where $C$ is the capacitance and $V$ is the potential - difference across the capacitor.

Step2: Convert capacitance to SI units

Given $C = 20\ \mu F=20\times10^{- 6}\ F$ and $V = 200\ V$.

Step3: Substitute values into the formula

$U=\frac{1}{2}\times(20\times10^{-6}\ F)\times(200\ V)^{2}$.
First, calculate $(200)^{2}=40000$. Then, $\frac{1}{2}\times20\times10^{-6}\times40000$.
$\frac{1}{2}\times20\times10^{-6}\times40000 = 10\times10^{-6}\times40000$.
$10\times10^{-6}\times40000=0.4\ J$.

Answer:

0.4 J