How to Find the Total Charge from its Density?

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Homework Help Overview

The problem involves calculating the total charge within a sphere of radius 0.2m, given a charge density that varies with position, specifically \(\frac{2}{\sqrt{x^{2}+y^{2}}}\) in units of c/m³. Participants are exploring how to integrate this density over the volume of the sphere to find the total charge.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest integrating the charge density over the volume of the sphere, while others discuss the need to express the integrand in spherical coordinates. There is also mention of the shell method and the potential advantages of using cylindrical-polar coordinates due to the symmetry of the density function.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the integration and emphasizing the importance of understanding the density function's behavior. There is no explicit consensus on a single method, as various approaches are being considered.

Contextual Notes

Participants note that the charge density is given in volume units, which raises questions about the appropriateness of certain equations that may imply surface charge density. There is also a focus on ensuring that the integration accounts for the variable nature of the density.

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Homework Statement



A ball whose its center on the origin, with radius of 0.2m, and contains a charge density [itex]\frac{2}{\sqrt{x^{2}+y^{2}}}c/m^{3}[/itex]


Homework Equations



[itex]D=\frac{\psi}{S}=\frac{\psi}{4\pi r^{2}}[/itex]

Where [itex]\psi=Q[/itex].

The Attempt at a Solution



The density is [itex]c/m^{3}[/itex], which is volume not surface.
Anyway, I've used the previous equation [itex]\frac{2}{\sqrt{x^{2}+y^{2}}}=\frac{Q}{4×\pi ×0.2^{2}} \rightarrow Q=\frac{2×4×\pi×0.2^{2}}{\sqrt{x^{2}+y^{2}}}[/itex]

Is that right?
 
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You have to integrate the density over the sphere volume.
 
You mean [itex]\int\limits_0^{2\pi} \int\limits_0^ \pi \int\limits_0^{0.2} \frac{2}{\sqrt{x^{2}+y^{2}}} r^{2} sin\theta dr d\theta d\phi[/itex]?
 
JasonHathaway said:
The density is [itex]c/m^{3}[/itex], which is volume not surface.
Anyway, I've used the previous equation [itex]\frac{2}{\sqrt{x^{2}+y^{2}}}=\frac{Q}{4×\pi ×0.2^{2}} \rightarrow Q=\frac{2×4×\pi×0.2^{2}}{\sqrt{x^{2}+y^{2}}}[/itex]
That formula would give you a charge per unit area and is only applicable when the density is uniform.

JasonHathaway said:
You mean [itex]\int\limits_0^{2\pi} \int\limits_0^ \pi \int\limits_0^{0.2} \frac{2}{\sqrt{x^{2}+y^{2}}} r^{2} sin\theta dr d\theta d\phi[/itex]?
Yes, ##Q = \int \rho\,\text{d}V## in general and only when ##\rho## is not dependent on V (i.e is uniform throughout the volume) does this simplify to ##Q = \rho \int \, \text{d}V = \rho V##. But you should express your integrand in spherical polar coordinates.
 
JasonHathaway said:
You mean [itex]\int\limits_0^{2\pi} \int\limits_0^ \pi \int\limits_0^{0.2} \frac{2}{\sqrt{x^{2}+y^{2}}} r^{2} sin\theta dr d\theta d\phi[/itex]?

yes, but express x2+y2 also in terms of spherical coordinates.

ehild
 
That's what I had in mind - yep.
The symmetry of the density function suggests cylindrical-polar may be better.
Maybe look at the shell method?

in general, the volume element ##\text{d}V## at position ##\vec r## has charge ##\text{d}q=\rho(\vec r)\text{d}V##

The total charge ##Q## in a volume ##V## is the sum of all those ##\text{d}q## elements: ##Q=\int_V\text{d}q##

Always start with a sketch - try to understand the density function: where is it a maximum, where a minimum, how does it change, etc?
 

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