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

Answer:

$a. 4.2057\times 10^-^1^2 F \ or 4.2057\ pF\\b. 2.7344V$

Explanation:

a.

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

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

The values are given as:$a=0.550mm(5.5\times 10^-^4m), \ b=4.00mm(4.0\times10^-^3m), \ L=15.0cm(0.150m)$

Substitute in our capacitance equation:

$C=\frac{2\times\pi \times 8.854\times 10^-^1^2 \times 0.15}{In(4.00/0.550)}\\=4.2057\times 10^-^1^2 F$

Hence the capacitance is $4.2057\times 10^-^1^2 F$

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,$C=4.2057\times 10^-^1^2 F$

We substitute $C$ in the pd equation:

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

Hence, the applied potential difference is 2.7344V