Calculating Capacitance & Inductance of Cylindrical Conductors

[sachi]

We have to find the capacitance and inductance of a pair of cylindrical conductors of radius r and at a distance d b away from each other. I get the following expressions

C = 2Pi*epsilon0/ln(b/a)
L = Mu0/2Pi *ln(b/a)

We now have a cylindrical conductor a distance d above an infinite conducting plane. We are supposed to get the following expressions using the method of images:

C = Pi*epsilon0/ln(2d/r)
L = Mu0/Pi * ln(2d/r)

I can understand why the b goes to 2 d, but I can't see why we miss the factor of 2 from the numerator of C and the denominator of L.

thanks very much

 

[Jhero]

for your help
Thank you for your question. it is important to understand the derivation of equations and how they are related to each other. In this case, the expressions for capacitance and inductance using the method of images are slightly different from the expressions for a pair of cylindrical conductors.

First, let's review the expressions for capacitance and inductance in a pair of cylindrical conductors. The capacitance is given by:

C = 2πε0/ln(b/a)

where ε0 is the permittivity of free space, b is the distance between the cylindrical conductors, and a is the radius of the conductors.

The inductance is given by:

L = μ0/2π * ln(b/a)

where μ0 is the permeability of free space.

Now, let's look at the method of images. In this method, we consider the system to be a cylindrical conductor placed above an infinite conducting plane. This creates an image of the conductor on the other side of the plane, which we can use to calculate the capacitance and inductance.

Using this method, the capacitance is given by:

C = πε0/ln(2d/r)

where d is the distance between the conductor and the plane, and r is the radius of the conductor.

Similarly, the inductance is given by:

L = μ0/π * ln(2d/r)

As you can see, the expressions for capacitance and inductance using the method of images have a factor of 1/2 compared to the expressions for a pair of cylindrical conductors. This is because the image conductor effectively doubles the distance between the conductors, hence the factor of 2 in the denominator.

I hope this helps to clarify why the expressions are slightly different. If you have any further questions, please don't hesitate to ask.