Question

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 18.5 pC. The inner cylinder has a radius of 0.300 mm, the outer one has a radius of 4.80 mm, and the length of each cylinder is 16.0 cm.

Part A. What is the capacitance?

Use 8.854×10⁻¹² F/m for the permittivity of free space.

Part B. What applied potential difference is necessary to produce these charges on the cylinders?

Answer 1

Part A

Capacitance of cylindrical capacitor is given by,

C = 2*pi**L/(ln(R_out/R_in))

given, = 8.854*10^−12 F/m = the permittivity of free space

R_out = 4.80 mm = 4.80*10^-3 m

R_in = 0.300 = 0.30*10^-3 m

L = length of cylinder = 16.0 cm = 0.16 m

So,

C = 2*pi*(8.854*10^-12)*0.16/(ln((4.80*10^-3)/(0.30*10^-3)))

C = 3.21*10^-12 F

C = 3.21 pF

Part B

Since, Q = C*V

given, Q = charge on capacitor = 18.5 pC = 18.5*10^-12 C

So, V = potential difference = Q/C

V = (18.5*10^-12)/(3.21*10^-12)

V = 5.76 volts