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?

Answer 1

E = 9.62*10^-8 J

Step-by-step explanation:

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

`E=(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=(2\pi \epsilon_o L)/(ln((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=(1)/(2)((2\pi \epsilon_o L)/(ln((r_2)/(r_1))))V^2\\\\E=(\pi \epsilon_o L)/(ln((r_2)/(r_1)))V^2\\\\E=(\pi (8.85*10^(-12)C^2/Nm^2)(150m))/(ln((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