# Homework Help: Parallel-plate air capacitor Problem

1. May 4, 2008

### TFM

[SOLVED] Capacitance Problem

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

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?

2. Relevant equations

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

3. 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

2. May 4, 2008

### Staff: Mentor

Check your cm/mm to m conversion.

3. May 4, 2008

### TFM

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

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

TFM

4. May 4, 2008

### Staff: Mentor

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?

5. May 4, 2008

### TFM

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

TFM

6. May 4, 2008

### Staff: Mentor

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

7. May 4, 2008

### TFM

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

8. May 5, 2008

### 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

9. May 5, 2008

### R A V E N

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?

10. May 5, 2008

### TFM

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

The Value I gotfrom the corresponding text book, 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

11. May 5, 2008

### R A V E N

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

12. May 5, 2008

### Vuldoraq

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

13. May 5, 2008

### TFM

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 doesnt work with 12V for this one.

TFM

14. May 5, 2008

### Vuldoraq

With the battery disconnected, and no new circuit made, the capacitor is unable to discharge. Therefore it retains the charge previously given to it.

15. May 5, 2008

### TFM

Indeed it is, Thanks

Thanks to all those who helped,

TFM