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

$$
V_{a b}=E d
$$

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

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

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

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

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