Parallel plate capacitor problem

In summary, the conversation discusses the construction of a capacitor using two square metal plates and two different dielectrics, with a distance of 0.37 mm between the plates. The capacitance of the device is calculated using the equation for a parallel plate capacitor, taking into account the area and the dielectric constants of the two materials. There is also a discussion about the proper use of the formula and the consideration of two capacitors in series. The final answer for the capacitance is 1.15 x 10^-8 farad.
  • #1
Punchlinegirl
224
0
A capacitor is constructed from 2 square metal plates. A dielectric k= 4.93 fills the upper half of the capacitor and a dielectric k=10 fills the lower half of the capacitor. Neglect edge effects. The plate are separated by a distance of 0.37 mm and the length of the plates is 27 cm. Calculate the capacitance C of the device. Answer in units of pF.
I used the equation for C of a parallel plate capacitor.
[tex] kE_o A/d + kE_o A/d [/tex]
Where the area= 3.7 x 10^-4 * .135 = 4.995 x 10^-5
So (4.93)(8.85 x 10^-12)(4.995 x 10^-5)/ 3.7 x 10^-4 + (10)(8.85 x 10^-12)(4.995 x 10^-5)/ (3.7 x 10^-4)
= 1.78 x 10^-11 F.
Converting to pF gave me 1.78 x 10^-23 which isn't right... help please?
 
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  • #3
I think d has to be different for each one, but I'm not sure how to do that.
 
  • #4
consider two capacitors in series each having distance d = half of 0.37 mm.The first has dielectric k=4.93,second has k=10.

Or you may use the following equation-

C=EA/[d/k + d'/k'] here d=d'=half of 0.37 mm
 
  • #5
So would I do,
(8.85 x 10^-12)(.27)/ (3.7 x 10^-4/4.93) + ((1.85 x 10^-4)/10) = 2.55 x 10^-8 ?
 
  • #6
Punchlinegirl said:
So would I do,
(8.85 x 10^-12)(.27)/ (3.7 x 10^-4/4.93) + ((1.85 x 10^-4)/10) = 2.55 x 10^-8 ?

area is (.27 m)^2;also d=1.85 x 10^-4 m.
 
  • #7
Ok so it would be:
(8.85 x 10^-12)(.0729) / ((1.85 x 10^-4/4.93) + (1.85 x 10^-4/10)) =1.15 x 10^-8
Am I doing this right now?
 
  • #9
Punchlinegirl said:
Ok so it would be:
(8.85 x 10^-12)(.0729) / ((1.85 x 10^-4/4.93) + (1.85 x 10^-4/10)) =1.15 x 10^-8
Am I doing this right now?
yes,this should be the answer in farad.

If you are not allowed to use the formula directly,then it would be better to consider two capacitors in series rather than to deduce the formula.
 

Related to Parallel plate capacitor problem

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of electrical component that stores energy in the form of an electric charge. It consists of two conductive plates separated by an insulating material, or dielectric. When a voltage is applied to the capacitor, one plate accumulates a positive charge and the other accumulates a negative charge, creating an electric field between the plates.

2. How do I calculate the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the equation 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. Alternatively, if the capacitor has a specific dielectric material between the plates, the equation can be modified to include the permittivity of that material.

3. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by the area of the plates, the distance between the plates, and the permittivity of the material between the plates. In general, increasing the plate area or decreasing the distance between the plates will increase the capacitance. Using a material with a higher permittivity will also increase the capacitance.

4. How does the charge on a parallel plate capacitor change with applied voltage?

The charge on a parallel plate capacitor is directly proportional to the applied voltage. This means that as the voltage increases, so does the charge stored on the capacitor. The relationship between charge and voltage is given by Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

5. What are some practical applications of parallel plate capacitors?

Parallel plate capacitors have many practical applications in electronics, such as in power supplies, filters, and energy storage devices. They are also commonly used in radio frequency circuits and as sensors in various electronic devices. Additionally, parallel plate capacitors are used in many types of scientific research and experiments, including in particle accelerators and in the study of electrostatics.

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