Problems of E-field of a Continuous Charge Distribution

AI Thread Summary
The discussion revolves around the derivation of the electric field (E-field) from a continuous charge distribution, specifically transitioning from a simple equation dE = ke(dq/r²) to a more complex form. Participants reference the relationship between surface charge density (σ) and volume charge density (ρ), noting that ρ can be expressed as Q/(πR² x dz) for a cylindrical volume. There is some confusion regarding whether the discussion pertains to a sphere or a cylinder, with clarification that an infinitesimally thin slice (dz) can be treated as a short cylinder. The conversation emphasizes the application of calculus in understanding these geometric relationships in charge distributions. Overall, the thread highlights the complexities involved in calculating electric fields from continuous charge distributions.
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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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