Calculating Electric Dipole Moment of Point Charges Along Z-Axis

In summary: I think the prime is actually wrong because I wanted to write a vector x' as argument.In summary, the conversation is about calculating the electric dipole moment of point charges along the z-axis with varying distances and charge distribution. The formula for the dipole moment is given and the process of integration to evaluate it is discussed. The questions raised include whether to use a prime in the arguments of the delta functions, how to further evaluate the integrals, and what to do with the x-vectors. The conclusion is reached that the prime notation is incorrect and the integration is only along the x axis.
  • #1
joschua
3
0
Hi

I want to calculate the electric dipole moment of point charges along the z-axis with distances a and with the charge distribution

[tex] \varrho (\vec{x}) = q \delta (\vec{x}) - 2q \delta (\vec{x} - \vec{a}) + q \delta (\vec{x} - 2 \vec{a}) [/tex]

and of course [tex]\vec{a} = a \vec{e}_{z} [/tex]

I did the following:

[tex] \vec{p} = \int \vec{x}' \varrho (\vec{x'}) d^{3}x' [/tex]

[tex]= q \int \vec{x}' \delta (\vec{x}) d^{3}x' - 2q \int \vec{x}' \delta (\vec{x} - \vec{a}) d^{3}x' + q \int \vec{x}' \delta (\vec{x} - 2 \vec{a}) d^{3}x' [/tex]

Now I have some questions:

1.) I guess I should write a prime in the arguments of the delta functions. Is this true? (The definition of my electric dipole moment is with prime, the given distribution without but that makes no sense? I should write a prime to all x vectors or no primes. correct?

2.) How to evaluate the integrals further? I know that the delta function is only one at the points of the charges and everywhere else zero but what to do with the x-vectors?

If this would be a normal integral I would do integration by parts, but this makes no sense here.

In general I know the relation that

[tex]\int f(x) \delta (x-a) dx = f(a)[/tex]

but here I have no Function f because x is a vector and I am in 3-d space.

I am confused. Please help me.

edit:

I wanted to post it in Classical Physics and not here. Wrong forum. Sorry... maybe a nice mentor will move it? :)
 
Last edited:
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  • #2
1. All integration variables are x', including in the delta.
2. The f(x) here is just f(x)=x. You are integrating only along the x axis.
 
  • #3
thanks, I got it
 

1. What is the formula for calculating electric dipole moment?

The formula for calculating electric dipole moment is p = qd, where p is the electric dipole moment, q is the magnitude of the point charges, and d is the distance between the two charges.

2. How do you calculate the distance between two point charges?

The distance between two point charges, also known as the separation distance, can be calculated using the distance formula: d = √(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2. This formula takes into account the coordinates of the two point charges in the x, y, and z axes.

3. What units are used for electric dipole moment?

The SI unit for electric dipole moment is Coulomb-meter (Cm). However, it can also be expressed in Debye (D), where 1 D = 3.33564 x 10^-30 Cm.

4. Can electric dipole moment be negative?

Yes, electric dipole moment can be negative. This indicates that the two point charges have opposite polarities, with one being positive and the other being negative.

5. How is electric dipole moment related to electric field?

The electric dipole moment is directly proportional to the electric field strength. This means that as the electric dipole moment increases, the electric field strength also increases. The relationship between the two can be expressed as E ∝ p/d^3, where E is the electric field strength, p is the electric dipole moment, and d is the distance between the two point charges.

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