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 11.5 pC. The inner cylinder has a radius of 0.550 mm, the outer one has a radius of 4.00 mm, and the length of each cylinder is 15.0 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?

Answer 1

`a. 4.2057* 10^-^1^2 F \ or 4.2057\ pF\\b. 2.7344V`

Step-by-step explanation:

a.

Given the permittivity constant to be
`8.854* 10^-^1^2 F/m`,The capacitance of a
`cylindrical \ capacitor` of length,
`L` is given by the equation:

`C=(2\pi \epsilon _o L)/(ln(b/a))` where
`b` is the radius of the outer cylinder and
`a` the radius of the inner cylinder.

The values are given as:
`a=0.550mm(5.5* 10^-^4m), \ b=4.00mm(4.0*10^-^3m), \ L=15.0cm(0.150m)`

Substitute in our capacitance equation:

`C=(2*\pi * 8.854* 10^-^1^2 * 0.15)/(In(4.00/0.550))\\=4.2057* 10^-^1^2 F`

Hence the capacitance is
`4.2057* 10^-^1^2 F`

b. The charge on the capacitance is related to the potential difference across it. The potential difference is expressed using the equation:

`Q=CV`,
`Q=11.5pC`

From a above, we already have our capacitance value,
`C=4.2057* 10^-^1^2 F`

We substitute
`C` in the pd equation:

`v=>(11.5)/(4.2057)\\=2.7344V`

Hence, the applied potential difference is 2.7344V