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• Notes

$\begin{array}{l}{\text { The plates of a parallel-plate capacitor are } 2.50 \mathrm{mm} \text { apart, }} \\ {\text { and each carries a charge of magnitude } 80.0 \mathrm{nC} \text { . The plates are in }} \\ {\text { vacuum. The electric field between the plates has a magnitude of }} \\ {4.00 \times 10^{6} \mathrm{V} / \mathrm{m} \text { . (a) What is the potential difference between }} \\ {\text { the plates? (b) What is the area of each plate? (c) What is the }} \\ {\text { capacitance? }}\end{array}$

$V_{a b}=E d$

$E=\frac{Q}{\varepsilon_{0} A}$

$C=\frac{Q}{V_{a b}}$

(a) $V_{a b}=E d=4 * 10^{6} * 2.5 \times 10^{-3}$

$∴V_{a b}=1 * 10^{4} \mathrm{V}$

(b) $E=\frac{Q}{\varepsilon_{0} A} \rightarrow A=\frac{Q}{E \varepsilon_0}$

$\rightarrow A=\frac{80*10^{-9}}{4 * 10^{6}+8.854 * 10^{-12}}$

$=2.26 * 10^{-3} \mathrm{m}^{2}$

$=22.6 \mathrm{cm}^{2}$

(c) $C=\frac{Q}{V_{ab}}=\frac{80 \times 10^{-9}}{1 * 10^{4}}$

$=8 * 10^{-12} F$

$=8 pf$

$\begin{array}{l}{\text { A cylindrical capacitor consists of a solid inner conduct- }} \\ {\text { ing core with radius } 0.250 \mathrm{cm} \text { , surrounded by an outer hollow con- }} \\ {\text { ducting tube. The two conductors are separated by air, and the }} \\ {\text { length of the cylinder is } 12.0 \mathrm{cm} \text { . The caparated by air, and the }} \\ {\text { Calculate the inner radius of the hollow tube. (b) When the capaci- - }} \\ {\text { tor is charged to } 125 \mathrm{V} \text { , what is the charge per unit length } \lambda \text { on the }} \\ {\text { capacitor? }}\end{array}$

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

$=\frac{2 \pi\left(8 \cdot 85 * 10^{-12}\right)(0.12)}{3.67 * 10^{-11}}=0.182$

$∴ r_{b}=r_ a e^{0.182} = (0. 25cm)e^{0.182 }=0.3cm$

(b) $C=\frac{Q}{V}, \lambda=\frac{Q}{L} \longrightarrow$$\lambda=\frac{Q}{L}=\frac{C V}{L}$

$∴ \lambda=\frac{C V}{L}= \frac{3.67 * 10^{-11} * 125}{0.12} =3.82 * 10^{-8} \mathrm{C} / \mathrm{m}$

$={38.2 n c / m}$

$\begin{array}{l}{\text { A spherical capacitor contains a charge of } 3.30 \mathrm{nC} \text { when }} \\ {\text { connected to a potential difference of } 220 \mathrm{V} \text { . If its plates are sepa- }} \\ {\text { rated by vacuum and the inner radius of the outer shell is } 4.00 \mathrm{cm} \text { , }} \\ {\text { calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) }} \\ {\text { the electric field just outside the surface of the inner sphere. }}\end{array}$

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

(a) $c=\frac{Q}{V}=\frac{3.3 * 10^{-9}}{2.2 * 10^{2}}$$=1.5 * 10^{-11} \mathrm{F}$

$=15 \mathrm{PF}$

$C=4 \pi \varepsilon_{0} \frac{r_{a} r_{b}}{r_{b}-r_{a}}$

(b) $C=15 P F, \quad r_{b}=4_{c m}$$\longrightarrow r_{a}=3.09 \mathrm{cm}$

(c) $E=\frac{1}{4 \pi{\varepsilon}_{0}} \frac{q}{r^{2}}$

$=\frac{9*10^9*3.3*10^-9}{(0.0304)2}=3.12*10^4 N/ C$