E-field in a cylinder with a hole parallel to its axis

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SUMMARY

The discussion focuses on calculating the electric field inside a long insulating cylinder with a cylindrical hole parallel to its axis. The cylinder has a uniform charge density ρ, while the hole has no charge. The electric field inside the cylinder is given by E1(r) = rρ/(2ε0) for r ≤ R1. The solution involves using Gauss' law and the principle of superposition to find the electric field contributions from both the charged cylinder and the hole, with the need to adjust for the coordinate system centered on the larger cylinder.

PREREQUISITES
  • Understanding of Gauss' law in electrostatics
  • Familiarity with cylindrical coordinate systems
  • Knowledge of electric field calculations for charged cylindrical distributions
  • Basic vector analysis in physics
NEXT STEPS
  • Study the application of Gauss' law for different charge distributions
  • Learn about electric field calculations in cylindrical coordinates
  • Explore the principle of superposition in electrostatics
  • Investigate the effects of charge density variations on electric fields
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Students studying electromagnetism, physics educators, and anyone involved in solving electrostatic problems related to charged cylindrical objects.

fayled
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Homework Statement


A long insulating cylinder of radius R1 has a cylindrical hole parallel to the axis of the cylinder. The radius of the hole is R2, and the distance from the centre of the cylinder to the centre of the hole is a. There is a uniform fixed charge per unit volume ρ throughout the cylinder, except in the hole where there is no charge. Using a cylindrical coordinate system with the z axis along the centre of the cylinder, evaluate the components of the electric field inside the cylinder and inside the hole (you may neglect edge effects at the ends of the cylinder).

Homework Equations


Gauss' law

The Attempt at a Solution


I know I need to use superposition. I would like to:
Calculate the electric field of the cylinder of radius R1 alone with no hole. I did so to get E1=0.5ρr for r≤R1.
Calculate the electric field of the cylinder of radius R2 alone with charge in it (not easy!)
Subtract the fields from the second calculation from the first.

The major problem is that my cylindrical coordinate system is centred on the large cylinder. How can I compute the electric field in and around the second cylinder with such a coordinate system? I want a cylindrical Gaussian surface around the cylinder of radius R2. I just can't see how to do it :/

Thanks in advance.
 
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fayled said:

Homework Statement


A long insulating cylinder of radius R1 has a cylindrical hole parallel to the axis of the cylinder. The radius of the hole is R2, and the distance from the centre of the cylinder to the centre of the hole is a. There is a uniform fixed charge per unit volume ρ throughout the cylinder, except in the hole where there is no charge. Using a cylindrical coordinate system with the z axis along the centre of the cylinder, evaluate the components of the electric field inside the cylinder and inside the hole (you may neglect edge effects at the ends of the cylinder).

Homework Equations


Gauss' law

The Attempt at a Solution


I know I need to use superposition. I would like to:
Calculate the electric field of the cylinder of radius R1 alone with no hole. I did so to get E1=0.5ρr for r≤R1.
Calculate the electric field of the cylinder of radius R2 alone with charge in it (not easy!)
Subtract the fields from the second calculation from the first.

The major problem is that my cylindrical coordinate system is centred on the large cylinder. How can I compute the electric field in and around the second cylinder with such a coordinate system? I want a cylindrical Gaussian surface around the cylinder of radius R2. I just can't see how to do it :/

Thanks in advance.
Yes, you can (and should) use superposition.

Don't forget that Electric Field is a vector quantity.


Your expression for the electric field due to a solid cylinder with no hole is in error.

Its magnitude is: E1(r) = rρ/(2ε0) . - - You left out the ε0 . This is for the electric field inside a uniform cylindrical charge distribution.

For the region of the hole, use a charge density of -ρ . Note that for the region outside the hole, you will the general expression for the electric field outside a uniform cylindrical charge distribution.


It may help to consider the following vectors:
Define vector ##\ \vec a = a\,\hat r \,,\ ## which is a vector from the z-axis to the axis of the cylindrical hole and is perpendicular to the z-axis.

Define vector ##\ \vec {r_1} = r_1\,\hat r \,,\ ## which is a vector from the z-axis to the point of interest and is perpendicular to the z-axis.

Define vector ##\ \vec {r_2} = r_2\,\hat r \,,\ ## which is a vector from the z-axis to the point of interest and is perpendicular to the z-axis.
 

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