Problems of E-field of a Continuous Charge Distribution

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Homework Help Overview

The discussion revolves around the electric field generated by a continuous charge distribution, specifically focusing on the transition from a basic differential equation to a more complex formulation as illustrated in a referenced image. Participants are exploring the relationships between surface charge density and volume charge density in the context of geometric shapes like disks and spheres.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand how to derive a more complex equation from the basic differential equation for electric field. Questions arise regarding the definitions and relationships between surface charge density (σ) and volume charge density (ρ), particularly in relation to different geometric configurations.

Discussion Status

Some participants express uncertainty about their understanding of the concepts, particularly in distinguishing between the geometries of spheres and cylinders. There is acknowledgment of the relationships between charge densities, and a few participants indicate that they find the reasoning behind the equations to be correct, while still seeking further clarification.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is also a focus on the implications of using calculus to analyze thin slices of geometric shapes, which is central to their reasoning.

mysci
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Warning: Template not used to format problem
dE = ke(dq/r2)
How to change to complicated equation in the following picture?
 

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mysci said:
σ=Q/ πR2 and
ρ=Q/ V
Then ρ=Q/ (πR2 x dz) in here?
Looks right.
 
haruspex said:
Looks right.
In fact, I don't understand although I can deduce it.
The volume of sphere is (4/3)πR3, (πR2 x dz) = base area x height = volume of cylinder
Whether I got wrong about the picture is talking about the cylinder not sphere?

Thanks.
 
mysci said:
In fact, I don't understand although I can deduce it.
The volume of sphere is (4/3)πR3, (πR2 x dz) = base area x height = volume of cylinder
Whether I got wrong about the picture is talking about the cylinder not sphere?

Thanks.
dz indicates an arbitrarily thin slice. Such a thin slice of a sphere is effectively a very short cylinder. This is the basis of calculus.
 

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