A cylindrical capacitor is made of two concentric cylinders. The inner cylinder has radius r1 = 4 mm, and the outer one a radius r2= 8 mm. T

Question

A cylindrical capacitor is made of two concentric cylinders. The inner cylinder has radius r1 = 4 mm, and the outer one a radius r2= 8 mm. The common length of the cylinders is L = 150 m. What is the potential energy stored in this capacitor when a potential difference V = 4 V is applied between the inner and outer cylinder?

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Diễm Thu 3 weeks 2021-08-23T23:04:46+00:00 1 Answers 0 views 0

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    2021-08-23T23:06:18+00:00

    Answer:

    E = 9.62*10^-8 J

    Explanation:

    The energy stored in a capacitor is given by the following formula:

    E=\frac{1}{2}CV^2     (1)

    E: energy stored

    C: capacitance

    V: potential difference of the capacitor = 4 V

    The capacitance for a concentric cylindrical capacitor is:

    C=\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}     (2)

    L: length of the capacitor = 150m

    r2: radius of the outer cylinder  = 8mm = 8*10^-3m

    r1: radius of the inner cylinder = 4mm = 4*10^-3m

    εo: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2

    You replace the expression (2) into the equation (1) and replace the values of all parameters:

    E=\frac{1}{2}(\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})})V^2\\\\E=\frac{\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}V^2\\\\E=\frac{\pi (8.85*10^{-12}C^2/Nm^2)(150m)}{ln(\frac{8*10^{-3}m}{4*10^{-3}m})}(4V)^2\\\\E=9.62*10^{-8}J

    The energy stored in the cylindrical capacitor is 9.62*10-8 J

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