# Homework Help: Parallel Plate Capacitors Being Separated

1. Sep 25, 2008

### Fizzicist

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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?

2. Sep 26, 2008

### alphysicist

Hi Fizzicist,

What numbers did you use in the formula? and what answer did you get?

3. Sep 26, 2008

### Clairefucious

$$A = \pi (0.1m)^2$$

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

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

and

$$Q = C \times 15V$$?

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

4. Sep 26, 2008

### alphysicist

Hi Clairefucious,

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

$$A= 10 \ {\rm cm}^2$$

(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. Sep 26, 2008

### Fizzicist

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