Parallel-plate air capacitor Problem

  • #1


[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

[tex]|C = \epsilon_r\epsilon_0\frac{A}{d}[/tex]

The Attempt at a Solution


[tex]\epsilon_r[/tex] = 1.00059
[tex]\epsilon_r = 8.85*10^{-12}[/tex]
A = 0.0016m
d = 0.0047m,

And inserting into the formula, I get:

[tex]|C = (1.00059)(8.85*10^{-12})frac{0.0016}{0.0047}[/tex]

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

Any uides where I have have gone wrong?


Answers and Replies

  • #2
Check your cm/mm to m conversion.
  • #3
4.7mm = 0.0047m,
16 cm squared = 0.0016 square meters

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

  • #4
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
I entered the equation into Mastering Physics, and it just said it was wrong?

  • #6
No idea how these things work in physics, in the chemistry I would tell watch significant digits...
  • #7
In MP, you can enter the formula and it calculates it for you, so I entered:


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?

  • #8
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:

[tex]C = \frac{Q}{V}[/tex]

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

So, Any help will be very much greatly appreciated,

  • #9
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 [tex]\varepsilon_r=1.00059[/tex],in text of the problem?
  • #10
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?

  • #11
Sorry,I didn`t saw that it is an air capacitor.So maybe there is an error in program,althought probably not,but...
  • #12
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[tex]^{2}[/tex], it's 0.0256m[tex]^{2}[/tex]. 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
Thaks for the info, Vuldoraq, Silly Mastering Physics :rolleyes:

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.

  • #14
Your welcome.

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
Indeed it is, Thanks

Thanks to all those who helped,


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