The radius of the inner conductor of a long coaxial cable is...

The radius of the inner conductor of a long coaxial cable is a. The inner radius of the outer conductor is b. The inner and outer conductors are at potential V0 and 0, respectively. Determine the electric potential and electric field strength in the insulating material?

Transcript

00:01 Hello, in the question we have given that the radius of inner conductor of a long coaxial cable is a, the inner radius of outer conductor is b.

00:13 So, if i draw the cross -section of this conductor, so it will be like this.

00:19 So, i am drawing a cross -section.

00:21 So, this radius they have given it as a and this radius is given as b.

00:29 So, they have given one boundary condition that the inner and the outer conductor are at potential.

00:39 So, this is at potential v naught and this is at the potential zero.

00:42 So, we have to determine the electric potential and the electric field strength.

00:48 So, now in order to do this, we will use laplace equation.

00:52 So, now only the radial component will survive over here.

00:56 So, del square v will be equal to zero.

00:59 So, this is the laplace equation.

01:01 So, now we will use the cylindrical coordinates over here.

01:04 So, del square in cylindrical coordinate is this.

01:07 So, del square v is equal to zero.

01:10 So, if i plug in the laplace equation, so this is the radial part which we are interested in.

01:15 So, we will do this one upon r.

01:18 So, we will take the radial part.

01:20 So, the first component over here.

01:22 So, and we will equate it to zero.

01:25 So, if i send this r over here, i will get r del v by del r plus r del v by del r.

01:32 So, now, so this is differentiation of some constant because it is equal to zero.

01:39 We know that differentiation of any constant is equal to zero.

01:42 So, i can say that this is a constant.

01:45 So, r del v by del r will give me some constant.

01:49 Let us call that constant as a.

01:51 So, sending it to the other side.

01:53 So, rearranging this equation, i will get dv is equal to a da by r.

01:59 So, integrating on both the sides, we will get v as equals to a ln of r because integration of 1 upon r dr is ln of r plus b.

02:10 Let us call this as equation star.

02:12 Then we have given that at r is equal to b.

02:16 See, at r is equal to b...

Question

The radius of the inner conductor of a long coaxial cable is a. The inner radius of the outer conductor is b. The inner and outer conductors are at potential V0 and 0, respectively. Determine the electric potential and electric field strength in the insulating material?

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