How did they integrate the charge for Voltage question

In summary: Assuming you have looked up this integral in tables of same, I think what you're missing is the assumptions about the variables and z that would make the problem trivial if only you knew what they are. Even if you have success using the methods others have suggested earlier, looking for assumptions is a useful exercise that can pay off later on... As an aside, it took me a while to get Mathematica to solve your integral. The evaluation hung up until I specified the assumptions required to give an unambiguous answer (The same one, even!). MMA's both interesting and frustrating that way. In order to be as general as possible, it makes few assumptions but provides the user with methods to specify
  • #1
kiwibird4
8
0
I have a question on finding potential for a disc of charge when it comes to the simple integration
upload_2016-2-2_0-18-39.png

how does the integral of R'dR'/etc have the R'dR' just disappear? i thought it would then be R^2/2
I am obviously missing something here
 
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  • #2
Hi kiwibird4. Please remember to use the formatting template for all questions posted in the homework section.

The integral in question is not simply of R'dR'. Note that the denominator contains ##\sqrt{z^2 + R'^2}##.
 
  • #3
kiwibird4 said:
I have a question on finding potential for a disc of charge when it comes to the simple integration
View attachment 95177
how does the integral of R'dR'/etc have the R'dR' just disappear? i thought it would then be R^2/2
I am obviously missing something here
u-substitution ##u = z^2+R'^2## is applied.
 
  • #4
gneill said:
Hi kiwibird4. Please remember to use the formatting template for all questions posted in the homework section.

The integral in question is not simply of R'dR'. Note that the denominator contains ##\sqrt{z^2 + R'^2}##.

I'm SORRYYYY :cry::cry::cry::cry::cry::cry::cry::cry::H:H:H:H:H:H:H:H
I even had to check the lil button that said -I have followed the homework format
:cry:
 
  • #5
kiwibird4 said:
I'm SORRYYYY :cry::cry::cry::cry::cry::cry::cry::cry::H:H:H:H:H:H:H:H
I even had to check the lil button that said -I have followed the homework format
:cry:
Somehow we must all struggle through this great tragedy and believe in the promise of a brighter tomorrow :smile:

Cheers.
 
  • #6
Assuming you have looked up this integral in tables of same, I think what you're missing is the assumptions about the variables and z that would make the problem trivial if only you knew what they are. Even if you have success using the methods others have suggested earlier, looking for assumptions is a useful exercise that can pay off later on...

As an aside, it took me a while to get Mathematica to solve your integral. The evaluation hung up until I specified the assumptions required to give an unambiguous answer (The same one, even!). MMA's both interesting and frustrating that way. In order to be as general as possible, it makes few assumptions but provides the user with methods to specify them.
 

Related to How did they integrate the charge for Voltage question

1. How is charge integrated to find voltage?

To integrate charge and find voltage, you must use the equation V = Q/C, where V is voltage, Q is charge, and C is capacitance. This equation takes into account the relationship between charge and voltage in a capacitor.

2. What is the purpose of integrating charge for voltage?

The integration of charge for voltage is important because it allows us to understand the relationship between these two quantities in a capacitor. It also helps us to calculate the voltage across a capacitor when given the charge and capacitance.

3. Can you explain the process of integrating charge for voltage in more detail?

Integrating charge for voltage involves using calculus to find the area under a curve on a charge vs. time graph. This area represents the total charge that has been accumulated on a capacitor over time, and when divided by the capacitance, gives us the voltage across the capacitor.

4. Are there any limitations to integrating charge for voltage?

Integrating charge for voltage assumes that the charge on the capacitor is constant, which may not always be the case in real-life situations. It also assumes that the capacitance is constant, which may not be true for all types of capacitors.

5. How is integrating charge for voltage different from differentiating voltage for charge?

While integrating charge for voltage involves finding the area under a curve, differentiating voltage for charge involves finding the slope of the curve. This means that integrating is the inverse operation of differentiating, and they are both used in different ways to understand the relationship between charge and voltage in a capacitor.

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