Parallel plate capacitor and insulating material

In summary, for a parallel-plate capacitor with plates of 10 cm^2 area separated by a 0.10mm layer of glass insulation with resistivity rho = 1.2x10^13 omega*m and dielectric constant k= 5.6, the time constant for discharging through its insulation is 590s. This shows that the dependence is not on the dimensions, but rather on the properties of the insulating material. This is evidenced by the fact that the RC value does not vary significantly with changes in cross sectional area.
  • #1
tarellan
5
0

Homework Statement



A parallel-plate capacitor has plates of 10 cm^2 area separated by a 0.10mm layer of glass insulation with resistivity rho = 1.2x10^13 omega*m and dielectric constant k= 5.6. Because of the finite resistivity, current can leak through the insulation.

How do I Show that it depends only on the properties of the insulating material and not on its dimensions?

Homework Equations



C= emf(A/d)


The Attempt at a Solution


I found the time constant for this capacitor to discharge through its insulation. T= 590s
 
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  • #2
tarellan said:

Homework Statement



A parallel-plate capacitor has plates of 10 cm^2 area separated by a 0.10mm layer of glass insulation with resistivity rho = 1.2x10^13 omega*m and dielectric constant k= 5.6. Because of the finite resistivity, current can leak through the insulation.

How do I Show that it depends only on the properties of the insulating material and not on its dimensions?

Homework Equations



C= emf(A/d)


The Attempt at a Solution


I found the time constant for this capacitor to discharge through its insulation. T= 590s

Consider the resistivity

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/resis.html#c2

Consider too the equation you have for capacitance.

Notice anything about how - when you combine the 2 - the RC has a tendency not to be dependent on the cross sectional area?
 
  • #3


I would first clarify the question and make sure I understand what is being asked. The question is asking how to show that the leakage current through the insulation of a parallel plate capacitor depends only on the properties of the insulating material and not on the dimensions of the capacitor.

To answer this question, I would first remind the student that the leakage current is directly related to the resistivity of the insulating material. This means that the higher the resistivity, the lower the leakage current will be. Therefore, the properties of the insulating material, such as resistivity, play a crucial role in determining the leakage current.

Next, I would explain that the dimensions of the capacitor, such as the area of the plates and the distance between them, do not directly affect the resistivity of the insulating material. The resistivity is an intrinsic property of the material and is not affected by the size or shape of the capacitor. Therefore, the dimensions of the capacitor do not play a role in determining the leakage current.

To further illustrate this point, I would use the formula for capacitance, C= emf(A/d), where emf is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. This formula shows that the capacitance of a parallel plate capacitor is directly proportional to the area of the plates and inversely proportional to the distance between them. However, the resistivity of the insulating material is not included in this formula, further emphasizing that it is not affected by the dimensions of the capacitor.

In conclusion, the leakage current through the insulation of a parallel plate capacitor depends only on the properties of the insulating material, such as resistivity, and not on the dimensions of the capacitor. This is because the resistivity is an intrinsic property of the material and is independent of the size or shape of the capacitor.
 

Related to Parallel plate capacitor and insulating material

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of electronic component that consists of two conductive plates separated by an insulating material. It is used to store and release electrical energy.

2. How does a parallel plate capacitor work?

When a voltage is applied across the two plates, an electric field is created between them. The insulating material, also known as the dielectric, helps to increase the capacitance of the capacitor, allowing it to store more charge.

3. What is the role of the insulating material in a parallel plate capacitor?

The insulating material serves as a barrier between the two plates, preventing the flow of current between them. It also helps to increase the capacitance of the capacitor by storing energy in the electric field.

4. What are some common materials used as dielectrics in parallel plate capacitors?

Some common materials used as dielectrics in parallel plate capacitors include air, paper, plastic, ceramic, and glass. The type of material chosen depends on the desired capacitance, voltage rating, and other factors.

5. How do you calculate the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the formula C = εA/d, where C is capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. Alternatively, it can also be calculated using the formula C = Q/V, where Q is the charge stored on the plates and V is the voltage applied across them.

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