Finding the electric field on a cylinder with a hole

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SUMMARY

The discussion focuses on calculating the electric field inside a cylindrical hole of a charged insulating cylinder with uniform volume charge density "ρ". The first part establishes that the electric field outside the cylinder is given by E = (R²ρ)/(2rε₀) for r > R. The second part confirms that the electric field inside the hole at the center is zero due to Gauss's Law. The third part utilizes the principle of superposition to find the electric field at a distance "b" from the center, resulting in E = (bρ)/(2ε₀). The direction of the electric field in the third part is confirmed to be along vector "b".

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and charge distributions
  • Knowledge of superposition principle in electrostatics
  • Basic calculus for deriving electric field equations
NEXT STEPS
  • Study the application of Gauss's Law in various geometries
  • Learn about electric field calculations for non-uniform charge distributions
  • Explore the concept of superposition in electrostatics
  • Investigate the effects of different charge densities on electric fields
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Students and educators in physics, particularly those focusing on electromagnetism, as well as anyone involved in solving electrostatics problems related to charged cylinders.

PhysicsRob
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Homework Statement



A very long, solid insulating cylinder with radius "R" has a cylindrical hole with radius "a" bored along its entire length. The solid material of the cylinder has a uniform volume charge density "ρ". In this problem, you will find the magnitude and the direction of the electric field "E" inside the hole

(1) Find the magnitude and direction of the electric field of a cylinder similar to the one described above, but without the cylindrical hole.

(2) Find the magnitude and direction of the electric field inside the cylindrical hole when the cylindrical hole is in the center of the charged cylinder

(3)Find the magnitude and direction of the electric field inside the cylindrical hole when the center of the hole is a distance "b" from the center of the charged cylinder.

Homework Equations



Gauss' Law

The Attempt at a Solution



I solved through the first part for the solid cylinder by using a Gaussian cylinder with a radius of "r", r>R>a. I got that \vec{E} = (R2ρ)/2rε0 along vector "r". For the second part I thought that the field inside would be zero since if you drew a Gaussian cylinder inside the hole, wouldn't it contain no charge and thus have no electric field? For the third part I know that you need to use the idea of superposition, but I'm not really sure.
 
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PhysicsRob said:
for the solid cylinder by using a Gaussian cylinder with a radius of "r", r>R>a. I got that \vec{E} = (R2ρ)/2rε0 along vector "r".
Doesn't seem right. Shouldn't it increase as r increases up to R?
For the second part I thought that the field inside would be zero since if you drew a Gaussian cylinder inside the hole, wouldn't it contain no charge and thus have no electric field?
No, that's only inside a conductor (with no charge in the cavity).
For the third part I know that you need to use the idea of superposition, but I'm not really sure.
I'd use that for b and c. What makes you doubtful?
 
Well for the first part I calculated that electric field for a "r" that was greater than "R", so shouldn't it be decreasing as "r" gets greater? Instead of doing this, should I calculate the electric field at the border of the cylinder?

For the second part, well then could you explain what else I could do to find the electric field inside the hole? I'm not sure what else I would do.

For the third part, I'm just not that sure about how to do it.
 
PhysicsRob said:
Well for the first part I calculated that electric field for a "r" that was greater than "R", so shouldn't it be decreasing as "r" gets greater? Instead of doing this, should I calculate the electric field at the border of the cylinder?

For the second part, well then could you explain what else I could do to find the electric field inside the hole? I'm not sure what else I would do.

For the third part, I'm just not that sure about how to do it.

You are actually more interested in the field inside of the cylinder. And, yes, Gauss's law and symmetry will tell you the field inside of the hole is zero when it's in the center. The superposition idea you want is that if you superimpose a cylinder of charge in the shape of the hole but with charge density -ρ on top of the cylinder without the hole you can create the same charge pattern as the cylinder with the hole.
 
Okay, so I'm pretty sure I have it now:

So for the first part I got E=rρ/2ε0 in the direction of vector "r"

For the second part I got that E=0

For the third part I got that E=bρ/2ε0 in the direction of vector "r"

Does that seem right? The only thing I'm not totally sure about now is the direction of the field in the third part. Would it be in the direction of "r" or "r-b"?
 
Oh wait, nevermind. The third part would be in the direction of just vector "b" correct? Since it doesn't depend on "r"?
 
PhysicsRob said:
Oh wait, nevermind. The third part would be in the direction of just vector "b" correct? Since it doesn't depend on "r"?

To do the third part you want to sum two fields. One due to the original cylinder and the other due to the negatively charge density cylinder you use to create the hole. Use superpostition!
 

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