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3 years ago

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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?

1 answer:

vladimir1956 [14]3 years ago

4 0

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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