Parallel Plate Capacitors Being Separated

In summary, the problem involves a parallel-plate air capacitor with plates 10 cm square and spaced 4.7 mm apart, connected to a 15 V battery. When the plates are pulled apart to a separation of 9.9 mm, the new capacitance and charge on each plate need to be calculated using the equations C = Q/V = (permittivity of freespace)(A/D) and Q = C * 15V. However, the incorrect value of 10 cm^2 for the plate area may be causing the solution to not work.
  • #1
Fizzicist
54
0

Homework Statement



A parallel-plate air capacitor is made by using two plates 10 cm square, spaced 4.7 mm apart. It is connected to a 15 V battery. If the plates remain connected to the battery and the plates are pulled apart to a separation of 9.9 mm.

What is the new capacitance? What is the new charge on each plate?

Homework Equations



C = Q/V = (permittivity of freespace)(A/D)

The Attempt at a Solution



I attempted to solve for the new capacitance by simply substituting in the new separation value and it isn't working. Why?
 
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  • #2
Hi Fizzicist,

Fizzicist said:

Homework Statement



A parallel-plate air capacitor is made by using two plates 10 cm square, spaced 4.7 mm apart. It is connected to a 15 V battery. If the plates remain connected to the battery and the plates are pulled apart to a separation of 9.9 mm.

What is the new capacitance? What is the new charge on each plate?

Homework Equations



C = Q/V = (permittivity of freespace)(A/D)

The Attempt at a Solution



I attempted to solve for the new capacitance by simply substituting in the new separation value and it isn't working. Why?

What numbers did you use in the formula? and what answer did you get?
 
  • #3
What about your solution doesn't work? Are you using

[tex]A = \pi (0.1m)^2[/tex]

[tex] \epsilon_{0} = 8.854 \times 10^{-12} F/m[/tex]

[tex] C = \frac{\epsilon _{0} A}{0.0099m}[/tex]

and

[tex] Q = C \times 15V [/tex]?

These are the equations you listed in your relevant equations section, and they should work!
 
  • #4
Hi Clairefucious,

Clairefucious said:
What about your solution doesn't work? Are you using

[tex]A = \pi (0.1m)^2[/tex]

I don't think this is correct here.

I was not certain about what the poster meant by "two plates 10 cm square". I interpreted that to mean that the area is

[tex]
A= 10 \ {\rm cm}^2
[/tex]

(and converting cm2 to m2 is a common place to make a mistake) but if the problem wasn't quoted directly it could mean a square with sides of length 10cm. Maybe the poster will clarify the problem.
 
  • #5
Yes, I used 10 cm^2. This might be the problem.
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two conductive plates separated by an insulating material, known as the dielectric. It is used to store electrical energy by creating an electric field between the plates.

2. How are parallel plate capacitors separated?

Parallel plate capacitors can be separated by increasing the distance between the two plates or by changing the properties of the dielectric material. This changes the capacitance of the capacitor, which affects its ability to store electrical energy.

3. What happens to the capacitance when parallel plate capacitors are separated?

When parallel plate capacitors are separated, the capacitance decreases. This is because increasing the distance between the plates or changing the dielectric material decreases the electric field between the plates, reducing the capacitor's ability to store electrical energy.

4. How does the separation of parallel plate capacitors affect their ability to store charge?

The separation of parallel plate capacitors affects their ability to store charge by changing the capacitance. As the plates are separated, the capacitance decreases, which means the capacitor can store less charge. This is because the electric field between the plates decreases, making it more difficult for the capacitor to hold onto the stored charge.

5. What factors can influence the separation of parallel plate capacitors?

The separation of parallel plate capacitors can be influenced by several factors, such as the distance between the plates, the properties of the dielectric material, and the voltage applied to the capacitor. Additionally, the shape and size of the plates can also affect the separation and therefore, the capacitance of the capacitor.

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