# Parallel Plate Capacitors Being Separated

## 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?

alphysicist
Homework Helper
Hi Fizzicist,

## 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?

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

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

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

and

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

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

alphysicist
Homework Helper
Hi Clairefucious,

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

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

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

(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.

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