A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius , and the outer one a radius r2 = 8.0 mm. The common length of the cylinders is L = 150 m. What is the potential energy stored in this capacitor when a potential difference 4.0 V is applied between the inner and outer cylinder?

Respuesta :

Answer:[tex]9.6\times 10^{-8} J[/tex]

Explanation:

The capacitance of cylindrical capacitor is

C=[tex]\frac{2\pi \varepsilon _0L}{\ln\frac{r_2}{r_1}}[/tex]

C=[tex]\frac{2\pi \times 8.854\times 10^{-12}\times 150}{\ln\frac{8}{4}}[/tex]

C=[tex]1.2\times 10^{-8} F[/tex]

potential Energy stored in capacitor=[tex]\frac{1}{2}CV^2[/tex]

=[tex]\frac{1}{2}\times 1.2\times 10^{-8}\times 4^2[/tex]

=[tex]9.6\times 10^{-8} J[/tex]

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