Parallel Plate Capacitor and distance

In summary, the question involves finding the necessary separation distance for a parallel-plate capacitor with a capacitance of 1 pF and two plates of area 3X10^-4 m^2. While the equation C = Eo(A/d) seems straightforward, a mistake in manipulating the equation causes the answer to come out incorrect. A more general formula and dimensional analysis can help in solving similar problems.
  • #1
Trista
33
0
I'm going through my book and one of the Example exercises has the following problem:
Two plates, each of area 3X10^-4 m^2 are used to construc a parallel-plate capacitor with capacitance 1 pF. (a) Find the necessary separation distance.

Seems pretty obvious right? C = Eo(A/d)
So, I figure I'm finding d = C/(Eo)(A)
but it doesn't come out right, so I must be missing something either in my math (not surprising) or the area. Do I need to do a calculation with the area? Like 2 X A? That doesn't work either, and it doesn't seem right to do it that way. Help, please??
 
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  • #2
Its just your manipulation which is letting you down. It should be;

[tex]C = \frac{E_{0}A}{d}[/tex]
[tex]C \cdot d = E_{0}\cdot {A}[/tex]
[tex]d = \frac{E_{0}A}{C}[/tex]

-Hoot:smile:
 
  • #3
Hootenanny said:
Its just your manipulation which is letting you down.


You are awesome! Thank you!

:approve:
 
  • #4
If you are assume the dielectric is air (k ~ 1) and [itex] \epsilon_o = 8.854 x 10^{-12}[/itex] F/m, you may get d = 2.656 mm. A more general formula for this type question may be found here.
In agreement with Hoot, i noticed the same fault in rearranging your equation. It is a good idea to do a couple of things to double check maths. One is to do the algebra step by step, the other is to do a quick dimensional analysis of the final equation. In your case, you have A left in the denominator. The only way that would work and have answer come out in meters is to have [itex]m^3[/itex] in the numerator, which you don't have.
 
  • #5
Wow. Thank you, that is most helpful. The link is terrific. Again, thank you!
 

Related to Parallel Plate Capacitor and distance

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy. It consists of two parallel conducting plates separated by a dielectric material.

2. How does the distance between the plates affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the distance between the plates. As the distance between the plates increases, the capacitance decreases and vice versa.

3. What is the formula for the capacitance of a parallel plate capacitor?

The formula for the capacitance of a parallel plate capacitor is C = ε0A/d, where C is the capacitance, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

4. How does the dielectric material between the plates affect the capacitance of a parallel plate capacitor?

The dielectric material between the plates affects the capacitance of a parallel plate capacitor by increasing it. This is because the dielectric material has a higher permittivity than air, which allows for more electric field lines to pass through, resulting in a higher capacitance.

5. What are some real-world applications of parallel plate capacitors?

Parallel plate capacitors have various applications in electronics, such as in power supplies, filters, and memory devices. They are also used in sensors, actuators, and as components in electrical circuits.

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