Solving Potential of Line Charge: Square 6m on a Side

[FatoonsBaby71]

Homework Statement

A uniform line charge of density pl = 1nC/m is arrange in the form of a square 6m on a side. (In the z plane) Find the potential at (0,0,5) m

Answer:35.6 V

Homework Equations

R = The distance from the line to the point (0,0,5)
dL = ax dx + ay dy + az dz

The Attempt at a Solution

Well any point on the uniform line charge can be selected for this problem so I picked (0,3,0)
There for the distance is R= Sqrt((5-0)^2+ (0-3)^2)) Then I know that the potential of a charge distribution is equal to the integration of the whole volume of (p dV)/(4pi*Eo the Permittivity of free space*R)

Therefore, the x components don't matter because at both points they are zero. So I figured I would just integrate in terms of y get an answer and then integrate in terms of z and get an answer then the sum of those would be the total potential. However it doesn't seem to be coming out to the right answer.

Integral [0 to 3] pl/4pi*E0*R dy = 4.6241
Integral [0 to 5] pl/4pi*E0*R dz = 7.7068

does anyone know what I am doing wrong??
thanks for your help

 

[jalalmalo]

I'm sorry I don't know the answer. I'm just wondering about the z-plane. what does that even mean. a palce has two axis

 

[turin]

A uniform line charge of density pl = 1nC/m is arrange in the form of a square 6m on a side. (In the z plane) Find the potential at (0,0,5) m

I think that your problem statement is missing some information. I assume that "the z plane" means the xy plane (i.e. the z=0 plane). I assume that the square of line charges is centered on the origin.

Well any point on the uniform line charge can be selected for this problem so I picked (0,3,0)

This problem is not symmetric under exchange of any two points on the square. There is some symmetry, but that's not it.

 

[FatoonsBaby71]

Yes, I was wrong the uniform line charge is in the x-y plane where z = 0... I apologize.

So I really can't just pick a point on the line charge...i thought it wouldn't matter because it is uniformly distributed. Do you have any idea on how I could attack this problem??

 

[turin]

So I really can't just pick a point on the line charge...i thought it wouldn't matter because it is uniformly distributed. Do you have any idea on how I could attack this problem??

The uniformity makes the integrand simpler than if the distribution were generally nonuniform. Your hints are: integration and symmetry. Oh, and you may have a formula in your book for the potential due to a finite length of line charge, so think about that as a possibly alternative approach.