# Cylindrical Symmetry - Gauss's Law

1. Feb 15, 2009

### nissan4l0

I am having trouble understanding the concept of cylindrical symmetry in an infinitely long line.

Please picture a finite Gaussian cylinder enclosing a portion of the length of the line, parallel to the line.

My book states that there cannot be any component of E perpendicular to the radius of the cylinder; the electric field lines can only be radial, that is, through the side walls, not through the end caps.

As I understand it, the line charge is composed of many individual point charges; each point charge has an electric field that spreads out in all directions radially, including perpendicular to the end caps. Why do we assume that there is no field at the end caps?

2. Feb 15, 2009

### Staff: Mentor

See the attached diagram. Consider the fields at P produced by two equidistant small sections of the line charge. The y-components add, the x-components cancel. You can pair up all the sections of the line charge like this, one section to the left of P and one section to the right of P. The x-components of the field for each pair cancel so the total x-component is zero, and the field must have only a y-component (radial).

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3. Feb 15, 2009

### nissan4l0

I understand how the surface of the cylinder along the line has the perpendicular component of E. What I don't understand, is, my book claims that the ends have no field component perpendicular to them. I understand that Ey does not give any flux to the ends, but what about the Ex component at the cap?

I have attached a picture.

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4. Feb 15, 2009

### Staff: Mentor

But there's no $E_x$ component at the cap. E has no component parallel to the line charge anywhere. (So long as we can treat the line as infinitely long, of course.)

In the diagram I drew, P can be anywhere, including on the end cap of the Gaussian surface.

5. Feb 15, 2009

### nissan4l0

jtbell, I have continued to analyze your picture, and I believe it has helped me understand now. Your image is also applicable to the end caps, since it is an infinite line, the charges found outside of the Gaussian surface also contribute to the field at the end caps. Thus, the net flux through the caps is zero.

Last edited: Feb 15, 2009
6. Feb 15, 2009

### nissan4l0

Thank you very much, jtbell. I now understand.

You also brought up the fact that the scenario changes if the line was not infinitely long, which also makes sense.

I appreciate it!