Help compute integral (electric potential)

In summary, the conversation discusses the calculation of an integral involving a norm and vectors in R^2. The problem arises in medical physics and involves finding the electric potential over a surface with a constant density of dipoles. The integral is typically calculated over the charge distribution, but in this case, it is over an arbitrary plane. The conversation includes a visual representation of the problem and attempts to clarify the integral and its limits of integration.
  • #1
Carl140
49
0

Homework Statement



How to calculate the following integral:

Integral( 1/ |x-y| 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 ((y1-x1)^2 + (y2-x2)^2)].

The problem is: I need to integrate this over any plane in R^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.
 
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  • #2
Can you tell me the original problem? Where is this integral coming from?
 
  • #3
gabbagabbahey said:
Can you tell me the original problem? Where is this integral coming from?

Sure, it's a problem which arises from medical physics. The idea is that we may assume that the electrical activity in a certain piece of tissue has a continuous distribution of dipoles with constant density but this tissue is "separated" by the plane which formed by the exterior wall of the tissue and the interior wall of it.

I know the formula for the electric potential in this case is equal to potential = p * Integral( dS / |x-y| ) 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
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
gabbagabbahey said:
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?

Dear gabbagabbaney, that's exactly the integral I need to calculate.
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/presentations/pdf/ScollanThesisDefense.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?
 
Last edited by a moderator:

1. What is an electric potential?

An electric potential is the amount of work needed to move a unit positive charge from one point to another in an electric field. It is measured in volts (V).

2. How is electric potential related to electric field?

Electric potential is related to electric field by the equation V = -∫E⋅dr, where V is the electric potential, E is the electric field, and dr is the infinitesimal displacement in the direction of the field.

3. What is the formula for computing electric potential due to a point charge?

The formula for computing electric potential due to a point charge is V = kQ/r, where k is the Coulomb's constant, Q is the magnitude of the point charge, and r is the distance from the point charge to the point where the potential is being calculated.

4. How do you compute electric potential for a continuous charge distribution?

To compute electric potential for a continuous charge distribution, you can use the formula V = ∫k(dQ)/r, where k is the Coulomb's constant, dQ is an infinitesimal charge element, and r is the distance from the charge element to the point where the potential is being calculated.

5. What is the unit for electric potential?

The unit for electric potential is volts (V), which is equivalent to joules per coulomb (J/C).

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