- #1
penguin007
- 77
- 0
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!
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!