Verifying Electric Field Approximation of a Capacitor

In summary: Your physics textbook should have such an example, or you can look up the topic online.In summary, a capacitor with circular plates of radius a, separated by a distance d (d<<a) and V(t)=V_{0}sin(wt), has an electric field between the plates that can be approximated by \vec{E}(t)\approx E_{0} sin(wt)\hat{e}_{z}. The expression for E_{0} is -\frac{q_{0}}{ \epsilon_{0} A}, and this approximation is adequate when d is much smaller than a and there is minimal fringing of the electric field outside of the capacitor edges. When d is not significantly smaller than a and
  • #1
rmfw
53
0

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
 
Last edited:
Physics news on Phys.org
  • #2
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?
 
  • #3
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] ?
 
  • #4
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?
 
  • #5
hm no..
 
  • #6
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...
 
  • #7
when d is not << a?

but where is the mathematics to show it?
 
  • #8
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.
 

1. What is the purpose of verifying the electric field approximation of a capacitor?

The purpose of verifying the electric field approximation of a capacitor is to ensure that the theoretical calculations and assumptions made about the electric field within the capacitor are accurate. This helps to validate the accuracy of the model and can provide insight into any discrepancies between the theoretical and experimental results.

2. How is the electric field approximation of a capacitor calculated?

The electric field approximation of a capacitor is calculated by dividing the charge on one of the capacitor plates by the area of the plate and the permittivity of the surrounding material. This results in a uniform electric field between the two plates.

3. What factors can affect the accuracy of the electric field approximation of a capacitor?

Factors that can affect the accuracy of the electric field approximation of a capacitor include the shape and size of the capacitor plates, the distance between the plates, the dielectric material between the plates, and any external influences such as nearby objects or other electrical fields.

4. How can the electric field approximation of a capacitor be verified experimentally?

The electric field approximation of a capacitor can be verified experimentally by measuring the electric field strength between the plates using an electric field meter. This measurement can then be compared to the theoretical calculation to determine the accuracy of the approximation.

5. What are some potential sources of error when verifying the electric field approximation of a capacitor?

Potential sources of error when verifying the electric field approximation of a capacitor include measurement errors, variations in the dielectric material used, and external factors such as interference from other electrical fields. It is important to carefully control and account for these variables in order to obtain accurate results.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
104
  • Introductory Physics Homework Help
Replies
6
Views
304
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
376
  • Introductory Physics Homework Help
Replies
1
Views
875
  • Introductory Physics Homework Help
Replies
12
Views
188
  • Introductory Physics Homework Help
Replies
26
Views
557
  • Introductory Physics Homework Help
Replies
4
Views
884
  • Introductory Physics Homework Help
Replies
27
Views
2K
Back
Top