Answer:
Part(a): The relative capacitance is [tex]\dfrac{C_{A}}{C_{B}} = 0.33[/tex]
Part(b): The relative energy stored is [tex]\dfrac{U_{A}}{U_{B}} = 0.33[/tex]
Part(c): The relative charge stored is [tex]\dfrac{Q_{A}}{Q_{B}} = 0.33[/tex]
Explanation:
We know the capacitance ([tex]C[/tex]) of a capacitor having charge ([tex]Q[/tex]) and subjected to a potential difference of ([tex]V[/tex]) is given by
[tex]C = \dfrac{Q}{V}[/tex]
Also, the energy ([tex]U[/tex]) stored by a capacitor can be written as
[tex]U = \dfrac{1}{2}C~V^{2}[/tex]
Let us assume that the inner radius of the Capacitor B, as shown in the figure, be [tex]\textbf{r_{i}^{B}}[/tex][tex]\bf{r_{i}^{B}}[/tex], the outer radius be [tex]\bf{r_{o}^{B}}[/tex], the inner radius of Capacitor A be [tex]\bf{r_{i}^{A}}[/tex] and the outer radius be [tex]\bf{r_{o}^{B}}[/tex].
Given in the problem,
[tex]&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}[/tex]
Now, the capacitance ([tex]C[/tex]) of a cylindrical capacitor is given by,
[tex]\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}[/tex]
where [tex]\epsilon_{o}[/tex] is the permittivity of the free space, [tex]L[/tex] is the length of the cylindrical capacitor.
Part(a):
The capacitance of capacitor A,
[tex]C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}[/tex]
and the capacitance of capacitor B,
[tex]C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}[/tex]
giving the relative capacitance of each capacitor to be
[tex]\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33[/tex]
Part(b):
Energy stored by capacitor A,
[tex]U_{A} = \dfrac{1}{2}~C_{A}~V^{2}[/tex]
Energy stored by capacitor B,
[tex]U_{B} = \dfrac{1}{2}~C_{B}~V^{2}[/tex]
giving the relative energy stored by each capacitor to be
[tex]\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33[/tex]
Part(c):
The charge stored by capacitor A,
[tex]Q_{A} = C_{A}~V[/tex]
The charge stored by capacitor B,
[tex]Q_{B} = C_{B}~V[/tex]
giving the relative charge stored by each capacitor to be
[tex]\dfrac{Q_{A}}{Q_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33[/tex]
