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 12.0 pC. The inner cylinder has a radius of 0.250 mm, the outer one has a radius of 5.00 mm, and the length of each cylinder is 22.0 cm.1.What is the capacitance? Use 8.854×10−12 F/m for the permittivity of free space.2.What applied potential difference is necessary to produce these charges on the cylinders?

[tex]C=\frac{2\pi \epsilon o L}{ln (b/a)}[/tex] where and b are the inner and outer radius of teh cylinder, respectively. L is length of the cylinder.

Finally we also kwn that C=Q/ΔV

then we have

ΔV =Q/C

ΔV = 12 * 10^-12/17.65*10^-12= 0.68V

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T13:59:12+02:00

The Capacitance and the Potential difference are mathematically given as

C=17.65 * 10^-12 C/V
dV= 0.68V

What are the Capacitance and the Potential difference?

Question Parameter(s):

the magnitude of the charge on each is 12.0 pC

The inner cylinder has a radius of 0.250 mm

The outer one has a radius of 5.00 mm

Generally, the equation for the Capacitance is mathematically given as

[tex]C=\frac{2\pi \epsilon* L}{ln (b/a)}[/tex]

Therefore

[tex]C=\frac{2\pi 8.854*10^{-12 }* 22*10^(-2)}{ln ( 5.00*10^{-3}/0.250*10^{-3})}[/tex]

C=17.65 * 10^-12 C/V

In conclusion,the potential difference

dV =Q/C

dV = 12 * 10^-12/17.65*10^-12

dV= 0.68V