How to Find the Total Charge from its Density?

In summary: You have to integrate the density over the sphere volume.You mean \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?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 = \r
  • #1
JasonHathaway
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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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  • #2
You have to integrate the density over the sphere volume.
 
  • #3
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]?
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
 

FAQ: How to Find the Total Charge from its Density?

What is the formula for finding the charge using its density?

The formula for finding the charge using its density is Q = ρV, where Q is the charge in coulombs, ρ is the density in kilograms per cubic meter, and V is the volume in cubic meters.

How do I convert units when using the density formula?

When using the density formula, you may need to convert units in order to use the correct values. For example, if the density is given in grams per cubic centimeter, you will need to convert it to kilograms per cubic meter in order to use the formula.

What is the relationship between charge and density?

The relationship between charge and density is that charge is directly proportional to density. This means that as the density increases, the charge also increases, and vice versa.

Can I use the density formula for any type of charge?

Yes, the density formula can be used for any type of charge, including positive and negative charges. It is a general formula that applies to all types of charges.

What are the units for charge when using the density formula?

The units for charge when using the density formula are coulombs (C). This is the standard unit for charge in the SI system of units.

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