Electrostatics-Taylor expansion

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In summary, the conversation discusses the form of the Taylor expansion of the potential near the origin O in a Cartesian coordinate system, as well as the stability of a particle in a given electrostatic field. The answer to the first question is given as V(M)=V0+b*x+c*y+d*z+e*x^2+f*y^2+g*z^2+h*x*y+i*x*z+j*y*z, but the reason for this is not understood. The second question is linked to Earnshaw's theorem, which states that a field cannot be stable in three dimensions if it is stable in two dimensions. This explains why the movement is stable in xOy but not in (Oz).
  • #1
penguin007
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Hi everyone

First, I’m sorry if this topic is not in the right place here (I did hesitate with introductory physics).

My questions are the following:

In a Cartesian coordinate system:
Consider that you have four charges q placed in (a,0,0);(-a,0,0);(0,a,0);(0,-a,0).
The electrostatic field in O(0,0,0) will be null(because of the symmetries). But what is the form of the Taylor expansion of the potential near the origin O?

The answer is V(M)=V0+b*x+c*y+d*z+e*x^2+f*y^2+g*z^2+h*x*y+i*x*z+j*y*z
But I can’t understand WHY?


Second question:

Consider the electrostatic field E(M)=(x,y,-2*z) (in a Cartesian coordinate system).
“If you put a particle of charge q in M, then if it’s movement is stable in xOy, then this movement won’t be stable in (Oz)” Again, I can’t understand why.

Any help would be welcome!
 
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  • #2
Hi penguin007, you've been here long enough that you should know you need to show some attempt at a solution, or at least explain where you are stuck!

penguin007 said:
In a Cartesian coordinate system:
Consider that you have four charges q placed in (a,0,0);(-a,0,0);(0,a,0);(0,-a,0).
The electrostatic field in O(0,0,0) will be null(because of the symmetries). But what is the form of the Taylor expansion of the potential near the origin O?

The answer is V(M)=V0+b*x+c*y+d*z+e*x^2+f*y^2+g*z^2+h*x*y+i*x*z+j*y*z
But I can’t understand WHY?

What is the expression for the potential due to a collection of point charges?


Second question:

Consider the electrostatic field E(M)=(x,y,-2*z) (in a Cartesian coordinate system).
“If you put a particle of charge q in M, then if it’s movement is stable in xOy, then this movement won’t be stable in (Oz)” Again, I can’t understand why.

Any help would be welcome!

Read the section of your textbbook on Earnshaw's theorem.
 
  • #3
Hi, gabbagabbahey, sorry if my questions were not clear...
actually I guessed that I had to do a Taylor expansion at the order 2 to get the first result but didn't know its expression (I got it:https://www.physicsforums.com/showthread.php?t=346001).
For my second question, I understand it is linked to Earnshaw's theorem, but I still don't get the idea:how can this field be stable in xoy and not in (oz)? Any explanation would be welcome...
 
  • #4
penguin007 said:
For my second question, I understand it is linked to Earnshaw's theorem, but I still don't get the idea:how can this field be stable in xoy and not in (oz)? Any explanation would be welcome...

Well, Earnshaw's theorem tells you it can't be stable in three dimensions...If its stable in two dimensions, it must therefor be unstable in the third dimension.
 
  • #5
Therefore, it could also be stabe in (xoz) and not in (oy)?
 
  • #6
Yes.
 

1. What is electrostatics-Taylor expansion?

Electrostatics-Taylor expansion is a mathematical technique used to approximate the electrostatic potential for a charge distribution in terms of a series of terms with increasing powers of the distance from the center of the distribution. It is based on the Taylor series expansion, which is a way to represent a function as an infinite sum of terms.

2. Why is electrostatics-Taylor expansion important in physics?

Electrostatics-Taylor expansion is important because it allows us to approximate the electrostatic potential for complex charge distributions, which is a fundamental concept in physics. It also helps in solving various problems related to electric fields and charges, making it a useful tool in many areas of physics, such as electromagnetism and quantum mechanics.

3. How is electrostatics-Taylor expansion calculated?

To calculate electrostatics-Taylor expansion, we use the formula for the Taylor series, which states that the value of a function at a certain point can be expressed as the sum of its derivatives at that point multiplied by the corresponding powers of the distance from the center. This process is repeated for each term in the series, resulting in an approximation of the electrostatic potential.

4. What are the limitations of electrostatics-Taylor expansion?

One limitation of electrostatics-Taylor expansion is that it assumes that the charge distribution is continuous, which may not always be the case in real-world scenarios. It also becomes less accurate as the distance from the center of the distribution increases, as the higher-order terms in the series become more significant.

5. How is electrostatics-Taylor expansion used in practical applications?

Electrostatics-Taylor expansion is commonly used in practical applications such as designing electronic circuits and analyzing the properties of materials. It is also used in particle accelerators to calculate the electric fields generated by charged particles, and in the study of atoms and molecules to understand their electronic structures.

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