Proper method of integration

In summary, the conversation discusses the calculation of charge enclosed within a non-conducting solid sphere with varying charge density. The correct equation for calculating charge enclosed is q_{enc}=\int_{\mathcal{V}} \rho_E dV, where \rho_E is the charge density and dV is the differential volume element. The incorrect equation (1) is derived from assuming a constant charge density, which is not the case in this scenario.
  • #1
orthovector
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Homework Statement


A non conducting solid sphere with radius [tex] r_1 [/tex] has charge density [tex] \rho_E = \rho_o \frac{r_1}^{r} [/tex]
what is the charge enclosed for [tex] 0 < r < r_1 [/tex] inside the non conducting sphere?

Homework Equations


[tex] \frac{q_{enc}}^{\frac{4}^{3}} \pi r^3}} = \rho_E = \frac{dq_{enc}}^{4 \pi r^2 dr} [/tex]

(1) [tex] \frac{4}^{3} [/tex] [tex] \pi r^3 \rho_E = q_{enc} [/tex]
[tex] \frac{4}^{3} [/tex] [tex] \pi r^3 \rho_o \frac{r_1}^{r} [/tex] [tex] = \frac{4}^{3} [/tex] [tex]\pi r^2 \rho_o r_1 = q_{enc}[/tex]

[tex] \frac{8}^{3} [/tex] [tex]\pi \rho_o r_1 r dr= dq_{enc}[/tex]
WHY CAN'T I TAKE THIS INTEGRAL TO FIND ENCLOSED CHARGE?
[tex] \int_{0}^{r} \frac{8}^{3} [/tex] [tex]\pi \rho_o r_1 r dr [/tex] = [tex] \int_{0}^{r} dq_{enc} = Q_{enc} [/tex]I KNOW I MUST put [tex] \rho_E = \rho_o \frac{r_1}^{r} [/tex] with [tex] dq_{enc} = 4 \pi r^2 dr [/tex] befofe i take the integral, but I'm not sure why (1) does not work.
 
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  • #2
orthovector said:
[tex] \frac{q_{enc}}^{\frac{4}^{3}} \pi r^3}} = \rho_E = \frac{dq_{enc}}^{4 \pi r^2 dr} [/tex]

Why would this be true?:confused:

How can [tex]\rho_E= \frac{q_{enc}}{\frac{4}{3} \pi r^3}[/tex] and [tex]\rho_E= \frac{dq_{enc}}{4 \pi r^2 dr}[/tex] both be true?

That would imply [tex]dq_{enc}=\frac{3q_{enc}}{r}[/tex] which would mean you have some with unit of charge on the LHS equal to something with units of charge over distance on the RHS...clearly one, or both of those equations is wrong!:smile:

(1) [tex] \frac{4}^{3} [/tex] [tex] \pi r^3 \rho_E = q_{enc} [/tex]

This would be true if [itex]\rho_E[/itex] was a constant; but it clearly isn't since it varies like 1/r.

Instead of messing around with all these jumbled equations, go back to the definition of volume charge density...what is that?

You should see that the charge enclosed by a volume [itex]\mathcal{V}[/itex] is always given by the equation [tex]q_{enc}=\int_{\mathcal{V}} \rho_E dV[/tex] where [itex]dV[/itex] is the differential volume element and [itex]\rho_E[/itex] is the charge density in that region.

In this case, [itex]\rho_E[/itex] is not constant throughout the volume and so you can't take it outside the integral. That means that [itex]q_{enc}\neq\rho_E *\text{Volume}[/itex].
 

1. What is the proper method of integration?

The proper method of integration is using techniques such as substitution, integration by parts, trigonometric substitution, partial fractions, and trigonometric identities to find the anti-derivative of a function.

2. When should I use the substitution method for integration?

The substitution method is useful when the integral contains a function within a function, and the inner function can be replaced with a new variable to simplify the integration.

3. How do I know when to use integration by parts?

Integration by parts is helpful when the integral contains a product of two functions, and one of the functions becomes simpler when differentiated multiple times.

4. What is the purpose of using partial fractions in integration?

Partial fractions are used to decompose a rational function into simpler fractions, which can then be integrated using other techniques.

5. Can trigonometric identities be used in integration?

Yes, trigonometric identities can be used to simplify integrals involving trigonometric functions, such as sine, cosine, and tangent.

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