- #1
FatoonsBaby71
- 14
- 0
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