V to get C. Calculating Max Capacitance of Coaxial Capacitor

In summary, for a coaxial capacitor with an inner radius of 5mm, length of 3 cm, and voltage rating of 2kV, the maximum capacitance can be found using the equation C = (2πε_rh)/ln(b/a), where ε_r is the dielectric constant (2.3) and h is the distance between the two conductors. The outer radius (b) can be found using the equation V = (ρ_s a/ε_r)[ln(a)-ln(b)] with V = 2kV. It is important to use the permittivity of free space times the relative permittivity, rather than just the relative permittivity, in the equations.
  • #1
seang
184
0

Homework Statement


Consider a coaxial capacitor. If the inner radius is 5mm, the length is 3 cm, and the voltage rating of the capacitor is 2kV, what is the maximum capacitance if the dielectric between the two conductors is 2.3, and E breakdown is 15MV/m.

Homework Equations


[tex]
E_r = \frac{\rho_s a}{\epsilon_r r}
[/tex]
a is the inner radius and r is the radius between the inner and outer conductors at whch we want to find the E field. Since the E field will strongest nearest to the inner conductor, I'm using:
[tex]
E_{r,breakdown} = \frac{\rho_s}{\epsilon_r}
[/tex]
This is because I want to find the surface charge density nearest to the inner conductor that will !begin to breakdown the dielectric. I figure we never even want to start to breakdown the dielectric. I hope this makes sense.
[tex]
V = \frac{\rho_s a}{\epsilon_r}[ln(a)-ln(b)]
[/tex]
with V = 2k I used this to find b (outer radius)
Finally,
[tex]
C = \frac{2\pi\epsilon_r h}{ln(b/a)}
[/tex]

The Attempt at a Solution


I derived all the above equations, and pretty much plugged in numbers, and I'm getting about 1.6 farads which seem wrong.

Actually, thinking about it. I don't see why I can't just use gauss' law to find the enclosed charge (with E = E breakdown) and then divide by 2k
 
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  • #2
I got it. If anyone is interested, the mistake I made was deriving the equations by simply replacing the permittivity of free space with the relative permittivity. In fact you are supposed to replace the permittivity of free space with the permittivity of free space times the relative permittivity.
 
  • #3
V to find the capacitance.

I would like to commend your effort in deriving the equations and attempting to solve the problem. However, there are a few things that need to be clarified.

Firstly, the equation E_r = \frac{\rho_s a}{\epsilon_r r} is not correct in this context. This equation is used to calculate the electric field at a distance r from a charged conductor with radius a. In this problem, we are dealing with a coaxial capacitor, where the electric field is constant between the two conductors.

Secondly, the equation V = \frac{\rho_s a}{\epsilon_r}[ln(a)-ln(b)] is also not correct. This equation is used to calculate the potential difference between two points on a charged conductor. In this problem, we are dealing with a capacitor, where the potential difference is given by V = Ed, where E is the electric field and d is the distance between the two conductors.

To solve this problem, we can use the capacitance equation C = \frac{\epsilon_r A}{d}, where A is the area of the plates and d is the distance between them. In this case, the area A can be calculated as A = \pi(b^2-a^2), where b is the outer radius and a is the inner radius. Using the given values, we get A = 0.000176 m^2.

Next, we need to calculate the electric field between the two conductors. This can be done using the equation E = \frac{\rho_s}{\epsilon_r}, where \rho_s is the surface charge density and \epsilon_r is the relative permittivity of the dielectric. In this case, we are given the breakdown electric field E = 15 MV/m, so we can calculate \rho_s as \rho_s = E\epsilon_r = 15\times10^6\times2.3 = 34.5\times10^6 C/m^2.

Finally, we can calculate the capacitance as C = \frac{\epsilon_r A}{d} = \frac{2.3\times8.85\times10^{-12}\times0.000176}{0.03} = 2.67\times10^{-12} F or 2.67 pF.

In conclusion, the maximum capacitance of the coaxial capacitor is 2.67 pF
 

1. What is a V to get C calculation?

A V to get C calculation is a mathematical formula used to determine the maximum capacitance of a coaxial capacitor. It takes into account the voltage and frequency of the capacitor to determine its maximum capacitance value.

2. What is a coaxial capacitor?

A coaxial capacitor is a type of capacitor that consists of two conductive cylinders separated by a dielectric material. It is used to store electrical energy and is commonly used in radio frequency applications.

3. How is the maximum capacitance of a coaxial capacitor calculated?

The maximum capacitance of a coaxial capacitor is calculated using the formula C = (2*π*ε*ε0*r1*r2)/ln(r2/r1), where C is the capacitance, ε is the dielectric constant, ε0 is the permittivity of free space, r1 is the inner radius of the inner cylinder, and r2 is the outer radius of the outer cylinder.

4. What are the factors that affect the maximum capacitance of a coaxial capacitor?

The maximum capacitance of a coaxial capacitor is affected by the type of dielectric material used, the dimensions of the conductive cylinders, and the frequency and voltage applied to the capacitor.

5. How is a V to get C calculation used in practical applications?

A V to get C calculation is used to determine the maximum capacitance of a coaxial capacitor, which is important in designing and selecting the appropriate capacitor for a specific application. It ensures that the capacitor can handle the required voltage and frequency without risking failure or damage.

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