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• Notes

$\begin{array}{l}{\text { A } 0.350 \text { -m-long cylindrical capacitor consists of a solid }} \\ {\text { conducting core with a radius of } 1.20 \mathrm{mm} \text { and an outer hollow }} \\ {\text { conducting tube with an inner radius of } 2.00 \mathrm{mm} \text { . The two conduc- }} \\ {\text { tors are separated by air and charged to a potential difference of }} \\ {6.00 \mathrm{V} \text { . Calculate (a) the charge per length for the capacitor; (b) }} \\ {\text { the total charge on the capacitor; (c) the capacitance; (d) the energy }} \\ {\text { stored in the capacitor when fully charged. }}\end{array}$

(a) $V=\frac{(\lambda)}{2 \pi \varepsilon_{0}} \ln \left(r_{b} / r_{a}\right)$$\longrightarrow\lambda=\frac{2\pi\varepsilon_0 V}{\ln (r_b/r_a)}$        $\frac{4}{2}=2$

$∴ \lambda=\frac{6}{2\left(9*10^{9}\right) \ln \left(\frac{2}{1.2}\right)} =6.53 * 10^{-10} \mathrm{C/m}$

(b) $Q=\lambda L=6.53*10^{-10} * 0.35=2.29 * 10^{-10} \mathrm{C}$

(c) $C=\frac{Q}{V}=\frac{2.29 * 10^{-10}}{6}=3.81 * 10^{-11} \mathrm{F}$

(d) $U=\frac{1}{2} C V^{2}=\frac{1}{2}\left(3.81*10^{-11}\right)(6)^{2}=6.85 * 10^{-10} \mathrm{J}$

$\begin{array}{l}{\text { A cylindrical air capacitor of length } 15.0 \mathrm{m} \text { stores }} \\ {3.20 \times 10^{-9} \mathrm{J} \text { of energy when the potential difference between }} \\ {\text { the two conductors is } 4.00 \mathrm{V} \text { . (a) Calculate the magnitude of the }} \\ {\text { charge on each conductor. (b) Calculate the ratio of the radii of the }} \\ {\text { inner and outer conductors. }}\end{array}$

$U=\frac{1}{2} Q V \longrightarrow Q$

$C=\frac{Q}{V}$

$\frac{c}{L}=\frac{2 \pi \varepsilon_{0}}{\ln \left(r_{b}/r_a\right)} \quad (24.4)$

(a) $U \cdot \frac{1}{2} Q V \rightarrow Q$$=\frac{2 u}{V}=\frac{2\left(3.2 * 10^{-9}\right)}{4}$$=1.6*10^{-9} \mathrm{C}$

(b) $\frac{c}{L}=\frac{2 \pi \varepsilon_{0}}{\ln \left(r_{b}/r_a\right)}$$\longrightarrow \frac{r_{b}}{r_{a}}$$=\exp (2 \pi \varepsilon_0 L/C)$

$\frac{r_{b}}{r a}=\exp \left(2 \pi \varepsilon_0 \frac{L V}{Q}\right)$$=\exp \left(2 \pi \varepsilon_{0}(15)(4)/1.6*10^{-9})\right.=8.05$