Hi All,

I am taking a prep-test and did my best to an answer this question but cant seem to get the same answer as was given in the test. I am hoping a member of the forum might be willing to provide some guidance. I will provide my calculations attempt to solve to show I tried first and am not just looking for an answer - I want to learn and the text book is big paper weight with words I dont fully comprehend yet....

The question: A parallel capcitor consists of two aluminum foil sheets of dimensions 3cm x 50cm seperated by a sheet of waxed paper 0.15mm thick. If the dielectric constant of the waxed paper is 4.52, and the dielectric strength is 50, its capacitance is? Answer given by prep-test is 4pf

My attempt at the solution is:
1) convert the area to meters and the distance to meters, divide the area by the distance
2) multiply the product of the above step by the constant epsilon (or 8.85 x 10^-12)
3) multiply the product of the above step by the dielectric constant of the waxed paper 4.52
4) The dielectric strength I dont believe has any bearing on the capacitance as I understand it (or does it?).
((.03m x .5m) / 0.00015m)) x εo = 885.4187817 x 4.52 = 4.0021nf

Any guidance would be awesome!

Thanks

 

Sorry I should add to clarify, my solutions answer is 4.0021nf, whereas the test gives an answer of 4pf

 

I get 4.0002 x 10^-9, which is 4nF

 

I get 4.0002 x 10^-9, which is 4pF

4 * 10^-9 is 4nF
And that's what I make it too.

 

I get 4.0002 x 10^-9, which is 4pF

Thanks for responding Dodgydave... can you let me know where I deviated from looking at my calculations? Am I forgetting to take something into account with my calculations?

This is driving me bananas...

 

I get 4.0002 x 10^-9, which is 4pF

I believe you are mistaken. 1E-9 is a nanofarad.
1E-12 is a picofarad.
I don't know if your math is correct, but your interpretation is wrong.

Edit: Albert got here first.

 

Oh wait, sorry read your reply wrong I guess.... you are saying it is 4nf right?

 

Yes. Your answer is correct and the test answer was incorrect, provided you gave us the same numbers and units as are shown in the test.

 

Yes. Your answer is correct and the test answer was incorrect, provided you gave us the same numbers and units as are shown in the test.

Thanks DickCappales! My sanity is sound once again!