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

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SUMMARY

The electric field at point P, located 6 cm from the central axis of a thin insulating rod with a charge density of λ = +5 nC/m, is calculated using Gauss's law. The charge density leads to a surface charge density of σ = 26.526 nC/m². The resulting electric field is determined to be 1499 N/C, which is half of the initial calculation of 2998 N/C due to the influence of the conducting cylindrical shell, which has zero electric field inside it. The factor of 2 arises from the properties of electric fields near conducting and non-conducting sheets.

PREREQUISITES
  • Understanding of Gauss's law and its application in electrostatics
  • Familiarity with electric field concepts related to charge distributions
  • Knowledge of the properties of conducting and non-conducting materials
  • Ability to perform calculations involving charge density and electric field strength
NEXT STEPS
  • Study the application of Gauss's law in cylindrical symmetry scenarios
  • Learn about the electric field due to infinite line charges
  • Explore the differences in electric fields created by conducting versus non-conducting sheets
  • Investigate the implications of charge distribution on electric field calculations
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields generated by charged objects and their interactions with 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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