Line Charge and Charged Cylindrical Shell (Gauss law)

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SUMMARY

The forum discussion focuses on calculating the electric field components at a point near an infinite line charge and a charged cylindrical shell using Gauss's Law. The linear charge density of the insulating shell is determined to be λ2 = -2.093 μC/m. The x-component of the electric field at point P, located 7.5 cm from the line charge, is calculated to be Ex(P) = 0 N/C, while the y-component Ey(P) requires further clarification due to discrepancies in calculations. Participants emphasize the importance of correctly applying Gauss's Law and combining the electric fields from both charge distributions.

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  • Understanding of Gauss's Law and its applications in electrostatics.
  • Familiarity with electric field calculations for line charges and cylindrical shells.
  • Knowledge of linear charge density and volume charge density concepts.
  • Proficiency in using the permittivity of free space (ε0) in calculations.
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  • Review the derivation of electric fields from line charges using Gauss's Law.
  • Study the superposition principle for electric fields from multiple charge distributions.
  • Explore the implications of charge density on electric field strength and direction.
  • Practice solving similar problems involving charged cylindrical shells and line charges.
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Sneakatone
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Homework Statement


An infinite line of charge with linear density λ1 = 6.2 μC/m is positioned along the axis of a thick insulating shell of inner radius a = 2.7 cm and outer radius b = 4.4 cm. The insulating shell is uniformly charged with a volume density of ρ = -552 μC/m3.1) What is λ2, the linear charge density of the insulating shell?
-2.093 μC/m
2) What is Ex(P), the value of the x-component of the electric field at point P, located a distance 7.5 cm along the y-axis from the line of charge?
0 N/C

3) What is Ey(P), the value of the y-component of the electric field at point P, located a distance 7.5 cm along the y-axis from the line of charge?

Homework Equations


The Attempt at a Solution



number 3 is the one I am stuck on. I used gauss law and ended with the equation E(x)=landa/(2pi (8.85*10^-12 * r) and got 503541.06 N/C but it seems to be wrong.
 
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Sneakatone said:
number 3 is the one I am stuck on. I used gauss law and ended with the equation E(x)=landa/(2pi (8.85*1-^-12 * r) and got 503541.06 N/C but it seems to be wrong.
I get nearly double that. Please post your working.
 
ah! you beat me to it haruspex
 
I used gauss law
integral(E*dA)=q_in/epsilon
LHS E∫dA RHS ∫(from 0 to h) (λ1*dz)/(ε0)
2pi*dx*h*E= (λ1*h)/ε0
E=λ/(2*∏*ε0*dR)
 
Yes, those are the equations, but I asked to see the working. can't tell where you're going wrong without that. you are adding the fields for the two charges, right?
 

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