Need Help?

Subscribe to Physics 2

Subscribe
  • Notes
  • Comments & Questions

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

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

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

 

No comments yet

Join the conversation

Join Notatee Today!