Finding an electric field at a point from a line of charge

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Homework Help Overview

The discussion revolves around calculating the electric field at a point near a thin insulating rod with a given charge density, which is placed inside a conducting cylindrical shell. The problem involves applying Gauss's law to determine the electric field at a specified distance from the central axis of the setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss's law and the implications of charge density on the electric field. There is a focus on the calculations leading to different results and the reasoning behind the factor of 2 that appears in the context of conducting and non-conducting sheets.

Discussion Status

Participants are actively exploring the calculations and concepts involved in the problem. Some have suggested revisiting the principles related to electric fields near conducting and non-conducting sheets, while others have pointed out potential errors in the original calculations. There is an ongoing examination of the assumptions made in the setup.

Contextual Notes

There is mention of previous calculations related to surface charge density and the need to consider the geometry of the Gaussian surface in relation to the electric field. The discussion also highlights the importance of understanding the behavior of electric fields in the presence of conductors.

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Homework Statement



A thin insulating rod with charge density lambda=+5nC/m is arranged inside a thin conducting cylindrical shell of radius r =3 cm. The rod and the shell are on the same axis, and infinite in length.
What is electric field at point p? 6cm from the central axis.

exam7.jpg


Homework Equations


Gauss's law : E*dA = Qenc/epsilon

The Attempt at a Solution


The question before this was actually to find the sigma.. which was 26.526nC/m^2. So I used sigma in the gauss's law. E*dA( area of cylindrical wall) = (sigma*A)/epsilon and got 2998. However, the answer was exactly the half of that(1499) ... Why?
 
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Just revise the content on field near a non conducting sheet and a conducting sheet. A factor of 2 comes there as there is no field on one side of a conducting sheet as field inside conductor has to be zero. Also i think the answer should be same that you would get in the absence of the infinite cylindrical shell.
 
Physiqs said:

Homework Statement



A thin insulating rod with charge density lambda=+5nC/m is arranged inside a thin conducting cylindrical shell of radius r =3 cm. The rod and the shell are on the same axis, and infinite in length.
What is electric field at point p? 6cm from the central axis.

exam7.jpg


Homework Equations


Gauss's law : E*dA = Qenc/epsilon

The Attempt at a Solution


The question before this was actually to find the sigma.. which was 26.526nC/m^2. So I used sigma in the gauss's law. E*dA( area of cylindrical wall) = (sigma*A)/epsilon and got 2998. However, the answer was exactly the half of that(1499) ... Why?

i think you are making an error in calculation...
the line of charge has given charge per unit length. and the Gaussian cylindrical surface (hypothetical) to calculate the flux through it is to be constructed at 6 cm radius so take the area of this surface times the electrical intensity and apply Gauss theorem.
 
very close to a non-conducting sheet of charge E = σ/ε₀ and for a conducting sheet of charge E' = σ/(2ε₀)
 
Last post was not correct. It may be read as:

very close to a non-conducting sheet of charge E = σ/(2ε₀) and for a conducting sheet of charge E' = σ/ε₀
 

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