How Do You Apply Gauss's Law to a Non-Uniformly Charged Non-Conducting Cylinder?

AI Thread Summary
The discussion revolves around applying Gauss's Law to a long, solid, non-conducting cylinder with a non-uniform volume density defined as ρ = A*r². For part a, the electric field is calculated at a point within the cylinder, while part b requires calculating the field outside the cylinder at a distance of 10 cm from the axis. Participants clarify that the Gaussian surface can be placed outside the cylinder, and the original radius indicates the extent of the charge distribution but does not restrict the Gaussian surface's location. The key point is that Gauss's Law can still be applied effectively even when the point of interest is outside the charged object. Understanding the implications of the cylinder's radius is crucial for accurate calculations.
lcam2
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1. Homework Statement
A long, solid, non-conducting cylinder of radius 8 cm has a non-uniform volume density, ρ, that is a function of the radial distance r from the axis of the cylinder. ρ = A*r2 where A is a constant of value 3 μC/m5.

Homework Equations



Gauss's law EA = Q/(epsilon)
Q(enclosed)= Pv
Volume of cylinder = 2*(PI)*r*L

The Attempt at a Solution


Part a can be done using gauss's law, and integrating Q(enclosed)= pv; since the volume density is not constant.
The problem is part b, where the point where I have to calculate the field is outside the Gaussian surface.
 
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lcam2 said:
The problem is part b, where the point where I have to calculate the field is outside the Gaussian surface.
That's a puzzling statement, since you put the Gaussian surface wherever you need to in order to find the field. Please state the complete part b question.
 
Im sorry i didn't notice it was missing,

(a) What is the magnitude of the electric field 6 cm from the axis of the cylinder?

(b) What is the magnitude of the electric field 10 cm from the axis of the cylinder?
 
lcam2 said:
(b) What is the magnitude of the electric field 10 cm from the axis of the cylinder?
OK, so what's the problem in applying Gauss's law to solve this part, just like part a?

Hint: Your Gaussian surface will have a radius of 10 cm.
 
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so, basically what you are saying is that the original radius of the cylinder does not matter, am I correct?
 
lcam2 said:
so, basically what you are saying is that the original radius of the cylinder does not matter, am I correct?
It matters in that it tells you the extent of the charge distribution. But it doesn't prevent you from having a Gaussian surface outside of the cylinder.
 
Ok, thank u!
 
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