Using Gauss' theorem ande exploiting the cylindrical symmetry of the system, show

[blueyellow]

Homework Statement

A wire of length L and negligible transverse dimensions, made of an insulating material, is placed on the x-axis between the origin and the point (L,0). The wire has a uniform line charge density lambda.

using Gauss' theorem and exploiting the cylindrical symmetry of the system, show that the electric field at the point (x,y)=(L/2,L/100) is

E=(100lambda)/(2pi epsilon0 L)

The Attempt at a Solution

integral (S) E.da=E 2 pi r l
Q/(epsilon 0)=(lambda l)/(epsilon0)

2pi r l E=lambda l/epsilon0

E=(1/2 pi r l) (lambda l/epsilon0)
=lambda/(2 pi epsilon0 r)

but I don't know how to proceed from here

 

[Doc Al]

Plug in for the given value of r.

 

[blueyellow]

oh right! so r is just L/100 because it the wire is along the x axis, so I don't have to do L-L/2?

 

[Doc Al]

oh right! so r is just L/100 because it the wire is along the x axis, so I don't have to do L-L/2?

Right. The point where you want to evaluate the field is at a distance of L/100 from the wire, so that's the value of r you need to use.

 

[rwisz]

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I would like to first clarify that Gauss' theorem, also known as Gauss' law, states that the electric flux through any closed surface is equal to the total enclosed charge divided by the permittivity of free space (epsilon0). This theorem is a fundamental concept in electromagnetism and is often used to calculate the electric field in various situations.

In this problem, we are dealing with a cylindrical symmetry, which means that the electric field is the same at all points equidistant from the axis of the cylinder. This allows us to simplify the problem by using a cylindrical Gaussian surface, which is a cylindrical surface with its axis coinciding with the wire.

Using Gauss' theorem, we can write:

integral (S) E.da = Q/epsilon0

Where Q is the total charge enclosed by the cylindrical surface and S is the surface area of the cylindrical surface. Since we are only interested in the electric field at the point (x,y)=(L/2,L/100), we can choose a cylindrical Gaussian surface with a radius of L/2 and a length of L/100. The surface area of this cylinder can be calculated as:

S = 2 pi r l = 2 pi (L/2)(L/100) = pi L^2/100

Now, the total charge enclosed by this Gaussian surface is simply the line charge density multiplied by the length of the cylinder, which is L. Therefore, we can write:

Q = lambda L

Substituting these values into Gauss' theorem, we get:

integral (S) E.da = (lambda L)/epsilon0

Since the electric field is constant on this cylindrical surface, we can take it out of the integral:

E * integral (S) da = (lambda L)/epsilon0

The integral on the left-hand side is simply the surface area of the Gaussian surface, which we have already calculated as pi L^2/100. Therefore, we can write:

E * (pi L^2/100) = (lambda L)/epsilon0

Solving for E, we get:

E = (lambda L)/(pi epsilon0 L^2/100) = (100lambda)/(pi epsilon0 L)

which is the same result as given in the problem statement. Therefore, we have shown that using Gauss' theorem and exploiting the cylindrical symmetry of the system, we can calculate the electric field at any point along the x-axis, including the