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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