Question

problem 17.52
700 v is applied to a 2900 - pf capacitor.
part a
how much energy is stored?
express your answer using three significant figures and include the appropriate units.
pe = value units
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Explanation:

Step1: Recall energy - storage formula

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

The capacitance $C = 2900\ pF=2900\times10^{- 12}\ F$. The potential - difference $V = 700\ V$.

Step3: Substitute values into the formula

$U=\frac{1}{2}\times(2900\times10^{-12}\ F)\times(700\ V)^{2}$.
First, calculate $(700)^{2}=490000$. Then, $\frac{1}{2}\times2900\times10^{-12}\times490000$.
$\frac{1}{2}\times2900\times490000\times10^{-12}= \frac{2900\times490000}{2}\times10^{-12}=710500000\times10^{-12}\ J$.
$710500000\times10^{-12}\ J = 7.11\times10^{-4}\ J$ (rounded to three significant figures).

Answer:

$7.11\times10^{-4}\ J$