Parallel Plate Capacitors Being Separated

AI Thread Summary
The discussion centers on calculating the new capacitance and charge of a parallel-plate capacitor when the plates are separated from 4.7 mm to 9.9 mm while connected to a 15 V battery. Participants highlight confusion regarding the area calculation, with one suggesting that "10 cm square" could mean either 10 cm² or a square with 10 cm sides. The correct area should be converted to meters for accurate calculations. The equations provided, including C = (permittivity of free space)(A/D) and Q = C × V, are confirmed to be applicable if the area is correctly interpreted. Clarification on the area is essential for solving the problem accurately.
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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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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?
 
What about your solution doesn't work? Are you using

A = \pi (0.1m)^2

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

C = \frac{\epsilon _{0} A}{0.0099m}

and

Q = C \times 15V?

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

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

A = \pi (0.1m)^2

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

<br /> A= 10 \ {\rm cm}^2<br />

(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.
 
Yes, I used 10 cm^2. This might be the problem.
 
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