What is the charge on a circular segment of a wide charged electrode?

starrynight24
Messages
3
Reaction score
0

Homework Statement



The electric field strength 5cm from a very wide charged electrode is 1000N/C. What is the charge (in nC) on a 1cm diameter circular segment of the electrode?

Homework Equations



E=Q/EoA

n=Q/A

3. The Attempt at a Solution [/b

I tried to figure out n, so that I can plug it into n=Q/A to figure out Q. But it didn't work.
 
Physics news on Phys.org
Since the electrode is so wide compared to how far away the electric field is measured at, we may treat the electrode as an infinite insulating plane of charge for which the electric field is given by Gauss's Law. It yields E\equiv\frac{\sigma}{\epsilon_0} where \sigma is the area charge density of the plate. This is essentially what you had. You're given the electric field. Just make sure that when you calculate the area that you use the radius since you are given the diameter.
 
What equation do I need to use to apply the area and the electric field to get the charge? n=Q/A?
 
I'm not sure what your n is supposed to be.. Apparently is the area charge density which is conventionally denoted as \sigma. You are given the electric field so you can find the sigma and you also know the area so you can find the charge.
 
Two symbols are being used for charge density here:

n = Q/A

\sigma = Q/A
 
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