Finding the electric field on a cylinder with a hole

In summary, we use Gaussian cylinders and Gauss's law to determine the magnitude and direction of the electric field inside a long, solid insulating cylinder with a cylindrical hole. The electric field is found to be zero when the hole is in the center of the cylinder and in the direction of "b" when the center of the hole is a distance "b" from the center of the cylinder. We also use the idea of superposition to determine the electric field inside the hole by creating a cylinder of charge with the opposite charge density and adding it to the original cylinder without the hole.
  • #1
PhysicsRob
8
0

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 [itex]\vec{E}[/itex] = (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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  • #2
PhysicsRob said:
for the solid cylinder by using a Gaussian cylinder with a radius of "r", r>R>a. I got that [itex]\vec{E}[/itex] = (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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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"?
 
  • #6
Oh wait, nevermind. The third part would be in the direction of just vector "b" correct? Since it doesn't depend on "r"?
 
  • #7
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!
 

1. What is the formula for finding the electric field on a cylinder with a hole?

The formula for finding the electric field on a cylinder with a hole is E = kλ/(2πεr), where E is the electric field, k is Coulomb's constant, λ is the charge per unit length on the cylinder, ε is the permittivity of the material, and r is the distance from the center of the cylinder.

2. How does the presence of a hole affect the electric field on a cylinder?

The presence of a hole on a cylinder changes the distribution of charge, resulting in a non-uniform electric field. The electric field is strongest near the edges of the hole and decreases as you move away from it.

3. Can the electric field on a cylinder with a hole be negative?

Yes, the electric field on a cylinder with a hole can be negative. This occurs when the direction of the electric field is opposite to the direction of the charge on the cylinder.

4. How does the radius of the hole affect the electric field on a cylinder?

The radius of the hole has a significant impact on the electric field on a cylinder. As the radius increases, the electric field strength decreases. This is because a larger hole results in a larger area for the electric field to spread out, reducing its strength.

5. Can the electric field on a cylinder with a hole be zero?

Yes, the electric field on a cylinder with a hole can be zero. This occurs when the distance from the center of the cylinder is equal to the radius of the hole. In this case, the electric field contributions from the cylinder and the hole cancel each other out, resulting in a net electric field of zero.

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