An air filled coaxial line (assumed to be very long) has inner conductor radius

Question

An air filled coaxial line (assumed to be very long) has inner conductor radius a=1 [cm] and the radius of the inner surface of the outer conductor is b=3 [cm], has a linear charge density of 1 [C/m] evenly distributed on the inner conductor surface.

(a) What should the charge density on the inner surface of the outer conductor so that the line is charge neutral?

(b) What is the E field at a point of radius r between the inner and outer conductors? What is its value for r=1.5 [cm] and direction?

(c) Determine the electrostatic potential difference between the inner and outer conductors.

Explanation / Answer

Using Gauss's law the charge on the outer surface (inner surface of the outer conductor) must be the same as the charge in the middle

So you have a cross section of length L:

The charge on the inside is just Q = L. This must exactly cancel the charge on the surface:

Q = -L = A
--> find A the lateral surface area of the inner surface (a cylinder):

A = 2rh --> r is the radius, b, and h is the length of the cross-section (height of the cylinder).

2bL = -L --> = -/(2b) = -1 / (2 * pi * 3x10^-2) = -5.305 µC/m².........................ans a)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.