Electric Field inside a charged ring

yeezyseason3
Messages
16
Reaction score
0

Homework Statement


Given a charged ring in 2-d, what is the e-field inside the ring?

Homework Equations


Epoint = kq/r^2

The Attempt at a Solution


This isn't a homework question, but more of a problem I keep running into whenever I think about it. I assumed it was 0. I came to this conclusion because the gradient of potential is electric field, and potential within a ring is constant, therefore electric field is 0.
 
Physics news on Phys.org
You can check by considering that the electric field is a vector... the field dead center is easy, you are wondering about off-center right?
 
Simon Bridge said:
You can check by considering that the electric field is a vector... the field dead center is easy, you are wondering about off-center right?
Yea I am, I tried integrating and it got really messy, I assumed I was doing something wrong and instead assumed that potential is constant and hence e field is 0.
 
You "assumed" the potential was constant?
Didn't you calculate it?

Trying to do the vector calculus is nasty - try exploiting the line of symmetry through the center of the ring and the point you want to find the field for ... if you integrate equal angles either side of that line for a short arc, you should be able to find another similar arc on the opposite side that will integrate to equal and opposite field.
 
yeezyseason3 said:
I tried integrating and it got really messy
If you show what you did, perhaps we can help ?
 
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...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top