# Parallel-plate air capacitor Problem

• TFM

#### TFM

[SOLVED] Capacitance Problem

## Homework Statement

A parallel-plate air capacitor is made by using two plates 16 cm square, spaced 4.7 mm apart. It is connected to a 12-Volt battery.

What is the capacitance?

## Homework Equations

$$|C = \epsilon_r\epsilon_0\frac{A}{d}$$

## The Attempt at a Solution

Using:

$$\epsilon_r$$ = 1.00059
$$\epsilon_r = 8.85*10^{-12}$$
A = 0.0016m
d = 0.0047m,

And inserting into the formula, I get:

$$|C = (1.00059)(8.85*10^{-12})frac{0.0016}{0.0047}$$

Which is 3.01*10^-12, which is apparently incorrect.

Any uides where I have have gone wrong?

TFM

Check your cm/mm to m conversion.

4.7mm = 0.0047m,
16 cm squared = 0.0016 square meters

so it can't be the mm/cm converions?

TFM

Sorry, missed that squared thingy, looked at 16 cm -> 0.0016 m and it didn't look right... That's what happen when you don't pay attention to units.

OTOH - why do you think 3 pF is wrong?

I entered the equation into Mastering Physics, and it just said it was wrong?

TFM

No idea how these things work in physics, in the chemistry I would tell watch significant digits...

In MP, you can enter the formula and it calculates it for you, so I entered:

$$1.00059*(8.85*10^{-12})\frac{0.0016}{0.0047}$$

And MP Calculated it to be 3.01*10^-12, I have also had 3.02 and 3.00 (Both *10^-12), and none of these answers are accepted!?

Any other ideas?

TFM

Any ideas will be greatlt appreciated since there are at least three more parts related to this question, and the next one is:

What is the charge on each plate?

To which I need the capcitance to use the formula:

$$C = \frac{Q}{V}$$

So without the correct capacitance, I can't finish the questions!

So, Any help will be very much greatly appreciated,

TFM

What is Mastering Physics,some software for technical computing?

And how exactly does it say that the result is wrong?

Where did you get that value of $$\varepsilon_r=1.00059$$,in text of the problem?

Mastering Physics is online Coursework Software - it basically is online homework:

The Value I gotfrom the corresponding textbook, Young and Freedman, 12th Edition, which accomapnies the Mastering Physics

All Mastering Physics Says is

"Try Again;"

So far I've tried 3.00, 3.01, 3.02, all x10^-12, none are correct?

TFM

Sorry,I didn`t saw that it is an air capacitor.So maybe there is an error in program,althought probably not,but...

Hey TFM,

I have the same assignment and I was having the same problem. It seems it is an error in the language used by MP. The area isn't 16cm$$^{2}$$, it's 0.0256m$$^{2}$$. The 16cm refers to the side length. Just to save you from entering the wrong answer too many times my value is (using your formulae);

4.8*10^-11 F

Thaks for the info, Vuldoraq, Silly Mastering Physics One small query further on,

"The battery is disconnected and then the plates are pulled apart to a separation of 9.4 mm.

What is the charge on each plate in this case?"

I have already calculated the Capacitance, but wasn't sure how to work this one out, Q=CV doesn't work with 12V for this one.

TFM