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

AI Thread Summary
The discussion focuses on applying Gauss's Law to a uniformly charged filament within a cardboard cylinder. The electric field at the surface of the cylinder is calculated to be approximately 51.382 kN/C, using the total charge of the filament and the dimensions of the cylinder. The total electric flux through the cylinder is determined to be 645 Nm²/C. It is emphasized that the Gaussian surface should match the length of the cylinder, which is 2.00 cm, rather than the entire length of the filament. The approximation of treating the filament as infinitely long is noted as a key consideration in this analysis.
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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