Verifying Electric Field Approximation of a Capacitor

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Homework Help Overview

The discussion revolves around verifying the electric field approximation of a capacitor with circular plates, focusing on the relationship between the voltage and the electric field in a time-varying scenario. The original poster presents a problem involving the electric field between capacitor plates, defined by their radius and separation, and seeks to understand the conditions under which the approximation holds.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the gradient of voltage to find the electric field and question the conditions for the approximation's validity. There is discussion about the relationship between the electric field, voltage, and plate separation, with hints about the units of electric field providing insight into the peak electric field expression.

Discussion Status

Some participants have identified the expression for the peak electric field and are considering when the approximation may become less accurate. There is an ongoing exploration of the mathematical justification for the conditions under which the approximation holds, particularly regarding the dimensions of the capacitor.

Contextual Notes

Participants note that the approximation may not hold when the separation distance is not significantly smaller than the plate radius, indicating a potential limitation in the application of the simple electric field equation.

rmfw
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Homework Statement


consider a capacitor with circular plates of radius a, separated by a distance d (d<<a) and [itex]V(t)=V_{0}sin(wt)[/itex]

a)Considering the z axis to be the capacitor axis, verify that the electric field between the plates is , in good approximation, given by [itex]\vec{E}(t)\approx E_{0} sin(wt)\hat{e}_{z}[/itex].

What is the expression of [itex]E_{0}[/itex] ?

In which conditions is adequate this approximation of [itex]\vec{E}(t)[/itex] ?

Homework Equations



[itex]\vec{E}= \nabla V = \hat{e}_{r} \frac{\partial V}{\partial r} + \hat{e}_{\varphi} \frac{\partial V}{\partial \varphi} + \hat{e}_{z} \frac{\partial V}{\partial z}[/itex]

[itex]q=C V[/itex]

The Attempt at a Solution



I tried to find the electric field trough the above gradient formula but V has no dependence on r, phi, or z.
I also tried trough the second equation and got [itex]\vec{E}(t)= -\frac{q_{0}}{ \epsilon_{0} A} sin(wt) \hat{e}_{z}[/itex] with [itex]E_{0}= -\frac{q_{0}}{ \epsilon_{0} A}[/itex] but then I can't explain why is it just an approximation and in which conditions is it a good aproximation
 
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rmfw said:

Homework Statement


consider a capacitor with circular plates of radius a, separated by a distance d (d<<a) and [itex]V(t)=V_{0}sin(wt)[/itex]

a)Considering the z axis to be the capacitor axis, verify that the electric field between the plates is , in good approximation, given by [itex]\vec{E}(t)\approx E_{0} sin(wt)\hat{e}_{z}[/itex].

What is the expression of [itex]E_{0}[/itex] ?

In which conditions is adequate this approximation of [itex]\vec{E}(t)[/itex] ?

Homework Equations



[itex]\vec{E}= \nabla V = \hat{e}_{r} \frac{\partial V}{\partial r} + \hat{e}_{\varphi} \frac{\partial V}{\partial \varphi} + \hat{e}_{z} \frac{\partial V}{\partial z}[/itex]

[itex]q=C V[/itex]

The Attempt at a Solution



I tried to find the electric field trough the above gradient formula but V has no dependence on r, phi, or z.
I also tried trough the second equation and got [itex]\vec{E}(t)= -\frac{q_{0}}{ \epsilon_{0} A} sin(wt) \hat{e}_{z}[/itex] with [itex]E_{0}= -\frac{q_{0}}{ \epsilon_{0} A}[/itex] but then I can't explain why is it just an approximation and in which conditions is it a good aproximation

Hint -- What are the units of the electric field E? Does that give you a hint for what the peak electric field will be in terms of the voltage V and plate separation d?
 
berkeman said:
Hint -- What are the units of the electric field E? Does that give you a hint for what the peak electric field will be in terms of the voltage V and plate separation d?

[itex]E_{0}=\frac{V_{0}}{d}[/itex] ?
 
rmfw said:
[itex]E_{0}=\frac{V_{0}}{d}[/itex] ?

Yes, electric field has units of Volts/meter.

Do you have thoughts for when this will become less accurate for some capacitor configurations?
 
hm no..
 
What is the traditional capacitor equation for the capacitance as a function of the Area of the capacitor and the spacing d?

Look up that equation on Hyperphysics or Wikipedia (look under capacitor), and there should be mention of when that equation is less accurate. That will give you a clue as to when (where?) the E field equation is not so accurate...
 
when d is not << a?

but where is the mathematics to show it?
 
rmfw said:
when d is not << a?

but where is the mathematics to show it?

That is correct. When you have significant E field that is "fringing" outside of the edges of the capacitor, you can't use that simple equation for the capacitance. Instead, you need to integrate and take the fringe field into account.
 

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