Verifying Electric Field Approximation of a Capacitor

AI Thread Summary
The discussion focuses on verifying the electric field approximation for a capacitor with circular plates, where the electric field is expressed as E(t) ≈ E₀ sin(wt) along the z-axis. The expression for E₀ is derived as E₀ = V₀/d, indicating that the electric field is proportional to the voltage and inversely proportional to the plate separation. The approximation is valid under the condition that the distance d between the plates is much smaller than the radius a of the plates (d << a). When this condition is not met, particularly when fringing effects occur at the edges of the capacitor, the simple approximation becomes less accurate. The discussion emphasizes the importance of understanding the geometry of the capacitor to apply the electric field equations correctly.
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Homework Statement


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

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

What is the expression of E_{0} ?

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

Homework Equations



\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}

q=C V

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 \vec{E}(t)= -\frac{q_{0}}{ \epsilon_{0} A} sin(wt) \hat{e}_{z} with E_{0}= -\frac{q_{0}}{ \epsilon_{0} A} 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 V(t)=V_{0}sin(wt)

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

What is the expression of E_{0} ?

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

Homework Equations



\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}

q=C V

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 \vec{E}(t)= -\frac{q_{0}}{ \epsilon_{0} A} sin(wt) \hat{e}_{z} with E_{0}= -\frac{q_{0}}{ \epsilon_{0} A} 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?

E_{0}=\frac{V_{0}}{d} ?
 
rmfw said:
E_{0}=\frac{V_{0}}{d} ?

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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