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$$
\begin{array}{l}{\text { A } 5.80-\mu \mathrm{F} \text { , parallel-plate, air capacitor has a plate separa- }} \\ {\text { tion of } 5.00 \mathrm{mm} \text { and is charged to a potential difference of } 400 \mathrm{V} \text { . }} \\ {\text { Calculate the energy density in the region between the plates, in }} \\ {\text { units of } \mathrm{J} / \mathrm{m}^{3} .}\end{array}
$$

$$
U=\frac{1}{2} \varepsilon_{0} E^{2}
$$

$$
E=\frac{V}{d} \longrightarrow V=E d
$$

$$
E=\frac{V}{d}=\frac{400 \mathrm{V}}{5 \times 16^{-3}}=8 * 10^{4} \mathrm{V} / \mathrm{m}
$$

$$
u=\frac{1}{2} \varepsilon_{0} E^{2}=\frac{1}{2}\left(8.854 * 10^{-12}\right)\left(8*10^{4}\right)^{2}
$$

\(∴ U=0.0283 \ \mathrm{J} / {m}^ 3 \)

An air capacitor is made from two flat parallel plates $${1.50 \mathrm{mm}$$  apart. The magnitude of charge on each plate is  $${0.0180 \mu \mathrm{C}$$  when the potential difference is $$ 200 \mathrm{V}$$ . (a) What is the capacitance? (b) What is the area of each plate? (c) What maxi- mum voltage can be applied without dielectric breakdown?  (Dielectric breakdown for air occurs at an electric-field strength of $${3.0 \times 10^{6} \mathrm{V} / \mathrm{m}$$  .) (d) When the charge is  $$\mathrm{O.O} 18 \mathrm{O} \mu \mathrm{C},$$  what total  energy is stored? 

$$
c=\frac{Q}{V_{a b}}
$$$$
, C=\frac{\varepsilon_0 A}{d}
$$$$,
V_{a b}=E d
$$

$$
\text { Stored Enegy }
$$ $$
=\frac{1}{2} Q V
$$

(a) $$
C=\frac{Q}{V}=\frac{0.018*10^{-6}}{200}
$$$$
=9*10^{-11} F=90 P F
$$

(b) $$
C=\frac{\varepsilon_{0} A}{d}
$$$$
\longrightarrow \quad A= \frac{cd}{\varepsilon_0}
$$$$
=\frac{9 * 10^{-11} * 1.5*10^{-3}}{8 \cdot 854 * 10^{-12}}=0.0152
$$

(c) $$
V=E d=3 * 10^{6} * 1.5 * 10^{-3}=4.5 * 10^{3} \mathrm{V}
$$

(d) $$
\text { Energy }=
$$$$
\frac{1}{2} Q V=
$$$$
\frac{1}{2}\left(0.018*10^{-6}\right)(200)=1.8 \times 10^{-6}j
$$

$$
=1.8 \mathrm{MJ}
$$

OR $$
\frac{Q^{2}}{2 c}=
$$$$
\frac{\left(0.018 * 10^{-6}\right)^{2}}{2 * 9 * 10^{-11}}
$$

$$
=1.8 \mathrm{MJ}
$$

$$
\begin{array}{l}{\text { For the capacitor net-work shown in Fig. } \mathrm{E} 24.31} \\ {\text { the potential difference across } a b \text { is } 220 \mathrm{V} \text { . Find (a) the total }} \\ {a b \text { is } 220 \mathrm{V} \text { . Find (a) the total charge stored in this network; }} \\ {\text { (b) the charge on each capaci-tor; (c) the total energy stored }} \\ {\text { in the network; (d) the energy stored in each capacitor; (e) }} \\ {\text { the potential difference across each capacitor. }}\end{array}
$$


 

$$
C_{e q}=c_{1}+c_{2}
$$$$,
c=\frac{Q}{v}
$$$$,
U=\frac{1}{2} c v^{2}
$$

(a) $$
c_{eq}=c_{1}+c_{2}=35+75=110 n f
$$

$$
Q_{total}=C_{eq} V
$$$$
=110*10^{-4} * 220=24.2 \mathrm{MC}
$$

(b) $$
V_{1}=V_{2}=V=220 \mathrm{V}
$$

$$
Q_{1}=C_{1} V=35 * 10^{-9}*220
$$$$
=7 \cdot 7 MC
$$    $$Q_{1}+Q_{2}=Q_{eq, tot}$$

$$
Q_{2}=C_{2} V=75 * 10^{-9}*220
$$$$
=16.5 \mathrm{MC}
$$    $$7.7+165=24.2$$

(c) $$
U_{\text { tot }}=\frac{1}{2}
$$$$
C_{eq}
$$$$
V^{2} = \frac{1}{2} *\left(110 * 10^{-9}\right) 220^{2}=2.66 \mathrm{mJ}
$$

(d) $$
U_{1}=\frac{1}{2} C_{1} V^{2}
$$$$
=\frac{1}{2}\left(35*10^{-9}\right)(220)^{2}=
$$$$
0.85 \mathrm{mJ}
$$        $$U_{1}+U_{2}=U_{tot}$$

$$
U_{2}=\frac{1}{2} C_{2} V^{2}
$$$$
=\frac{1}{2}\left(75 * 10^{-4}\right)(220)^{2}
$$$$
=1.81 \mathrm{mJ}
$$        $$0.85+1.81=2.66 m j$$

(e) $$
V=220 \mathrm{V}
$$
for each capacitor

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