How Does Gauss's Law Apply to a Charged Filament Inside a Cardboard Cylinder?

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Homework Help Overview

The discussion revolves around applying Gauss's Law to a uniformly charged filament situated within a cardboard cylinder. The problem involves calculating the electric field at the surface of the cylinder and the total electric flux through it, given specific dimensions and charge values.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of using the total charge of the filament and the length of the Gaussian surface. There is a focus on whether the length of the cylinder should match the length of the filament for accurate calculations.

Discussion Status

Some participants have provided guidance on the appropriateness of the length used in calculations and the assumptions regarding the electric field's behavior near the ends of the filament. Multiple interpretations of the Gaussian surface length are being considered, with acknowledgment of the approximation of the filament as infinitely long.

Contextual Notes

There is a mention of the problem's constraints regarding the length of the Gaussian surface, as well as the implications of approximating the filament as infinitely long due to its relative size compared to the cylinder.

JosephK
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Homework Statement


A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 \muC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder.


Homework Equations



Gauss's Law
Area of a cylinder
\Phi=EA

The Attempt at a Solution



a) Flux is zero at the ends. The area of the gaussian surface is 2\pirL
EA = \frac{q_en}{\epsilon_0}

The charge enclosed is the total charge of the 7.00 m rod which is given.

E = \frac{Q}{\epsilon_0 2\pi rL}

The length of the cylinder does not cover the whole rod. Since the electric field is uniform, the electric field at the surface of a larger cylinder is the same and the charge enclosed is proportional to the length of the cylinder.

E = \frac{Q}{2 \pi rL \epsilon_0}

plugging in E = 51.382k N/C

B)
Flux is
\Phi= EA
=E(2\pirl
= 645 \frac{Nm^2}{C}
 
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Did you take Q to be the whole charge of 2.00 \mu C? You have to use the amount of charge inside your Gaussian surface.
 
Yes, but also I used L as the length of the rod. Is it OK? Thank you.
 
Last edited:
Yes, it is OK if you used L for the length of your Gaussian surface.

I think I should mention this here even thought it doesn't change the numerical answer. It seems that they want you to have your Gaussian surface 2.00 cm long and not 7.00 m long. That is because 7.00 m is much larger than 2.00 cm, so the filament can be thought of as one of infinite length. It says in the beginning "Using reasonable approximations...". The approximation you make here is that the electric field vector is radially outward from the filament. That is only true for a line charge of infinite length. Near the ends of the filament the electric field vector wouldn't point radially outward.
 
Thank you for letting me know.
 

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