0
$\begingroup$

I am struggling between connecting a simulation with an analytical equation.

One could calculate the electric field due to a ring with Charge Q, at a certain location on the x-axis using a formula described here.

https://medium.com/swlh/electric-field-due-to-a-uniformly-charged-ring-6b0adcbf7b8d

There are other sources as well for the formula. However in the formula you use a total charge "Q" on the ring. While I am trying to do a simulation in (StarCCM+), to calculate the Electric Field of a charged ring, the simulation is working, however the boundary condition, I could provide is only "Electric Potential", which is also the value I could get from real life conditions.

In real conditions, I am charging a ring by a certain voltage. And it is +/- with a certain frequency.

So how would I convert this voltage (electric potential) value I have to a Charge value on the ring? So I could compare my analytical results with simulation?

$\endgroup$
3
  • $\begingroup$ The link that you have provided does not work. $\endgroup$
    – Farcher
    Commented Oct 28, 2022 at 7:12
  • $\begingroup$ Going via the potential route as in this video might be easier although you will need to use Lagrange polynomials and then you should consider what happens if the conductor has a finite diameter. $\endgroup$
    – Farcher
    Commented Oct 28, 2022 at 9:01
  • $\begingroup$ How do you define the ring in the simulator? You define it as a circular conductor with finite conductivity? Or you define it as a perfect conductor? $\endgroup$ Commented Oct 29, 2022 at 5:31

1 Answer 1

0
$\begingroup$

It shouldn't be a problem at all. You will have the result different just by a scale. The voltage on the ring depends on the capacitance of the system through which it was charged: $$C = \frac{Q}{V}$$ where V is the ring potential, referenced to the ground of the charging system. You can write the electric field equation in the provided link in terms of $CV$, and it will be just a scaled version of the one in $Q$. Using a system with higher capacitance gives you the same amount of total charge $Q$ but at lower ring potential, and so on.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.