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

  • Thread starter Thread starter FatoonsBaby71
  • Start date Start date
  • Tags Tags
    Potentials
FatoonsBaby71
Messages
14
Reaction score
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
 
Physics news on Phys.org
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
 
FatoonsBaby71 said:
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.

FatoonsBaby71 said:
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.
 
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??
 
FatoonsBaby71 said:
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.
 

Thread 'Please tell me where I am going wrong in this integral'

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
View full post »

Thread 'Direction of the magnetic field of an oscillating charge'

Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
View full post »

Thread 'Initial conditions for orbits around a wormhole'

Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
View full post »

Similar threads

Potential of a charged ring in terms of Legendre polynomials
Replies
16
Views
4K
Electrostatic Potential Energy of a Sphere/Shell of Charge
Replies
2
Views
4K
The potential of a sphere with opposite hemisphere charge densities
Replies
6
Views
3K
Potential inside a cylindrical shell with a line charge
Replies
6
Views
4K
Dipole moment of given charge distribution
Replies
19
Views
2K
Potential inside a uniformly charged solid sphere
Replies
42
Views
6K
Equilibrium circular ring of uniform charge with point charge
Replies
7
Views
2K
Potential and charge on a plane
Replies
1
Views
2K
What Potential Should Be Used for Energy Stored in a Charged Sphere?
Replies
1
Views
1K
Finding the electrostatic potential of a square sheet.
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top