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Help compute integral (electric potential) 
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#1
May2009, 05:31 PM

P: 49

1. The problem statement, all variables and given/known data
How to calculate the following integral: Integral( 1/ xy dS) where x and y are vectors in R^2,  represents norm. So say x= (x1,x2), y=(y1,y2). Then the integral is [ dS/ sqrt ((y1x1)^2 + (y2x2)^2)]. The problem is: I need to integrate this over any plane in R^3. 3. The attempt at a solution So I assume the plane is of the form ax+by+cz+d =0. Then I really have no clue how to proceed. 


#2
May2009, 05:46 PM

HW Helper
P: 5,004

Can you tell me the original problem? Where is this integral coming from?



#3
May2009, 11:42 PM

P: 49

I know the formula for the electric potential in this case is equal to potential = p * Integral( dS / xy ) where   represents norm and p density (since its constant we can take it out of the integral). Hence the problem reduces to calculate this integral but over an arbitrary plane (which is the plane I explained above). But I'm very confused on which limits of integration should I choose and how would the area element should look. If someone could please explain this I'd really appreciate it. 


#4
May2009, 11:59 PM

HW Helper
P: 5,004

Help compute integral (electric potential)
Typically when one calculates electric potentials, it involves integrating over the charge distribution, and for a surface with a uniform charge distribution, you encounter an integral of the form:
[tex]\int_{\mathcal{S}} \frac{d^2x'}{\vec{x}\vec{x'}}[/tex] Where [itex]\vec{x}[/itex] is the vector from the origin (in whatever coordinate system you choose) to the field point (the point at which you want to determine the potential), [itex]\vec{x'}[/itex] is the vector from the origin to a point on the surface containing the charge distribution, and the integration is over the source points. The way you describe your integral makes no sense to me. You seem to be saying that [itex]\vec{x}[/itex] and [itex]\vec{y}[/itex] are vectors in some plane, and you are integrating over an entirely different plane. Is this really what you mean? 


#5
May2509, 06:43 PM

P: 49

Here S is a surface which is bounded by two planes. Here's a picture which illustrates the situation: http://www.ccbm.jhu.edu/doc/presenta...sisDefense.pdf Page 41, picture D. You can see the two planes I was referring two. Here "S" is the surface which is exactly in the middle. How can you calculate that integral? 


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