Capacitance potential

  • Thread starter Dx
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Dx
[SOLVED] Capacitance potential

Hello,

I have 2 questions here to ask for help with?

1) a .2F cap is desired. What area must the pplates have if they are to be seperated by a 2.2mm air gap?

My formula i used with substition was C=e_o (A/d) with e_o = 8.85x10^-12.

I rounded my answer it was 4.5 so rounded it to 5x10^7m^2 is this correct?


2)What is the potential at a distance of 5 x 10^-10 m from the nucleus of charge +60e?
I donno if my formula is correct? ids it V=Ed?

I donno how to solve for this, can you help point me that way.
Thanks!
Dx :wink:
 

Answers and Replies

  • #2
Tom Mattson
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Originally posted by Dx
1) a .2F cap is desired. What area must the pplates have if they are to be seperated by a 2.2mm air gap?

My formula i used with substition was C=e_o (A/d) with e_o = 8.85x10^-12.

I rounded my answer it was 4.5 so rounded it to 5x10^7m^2 is this correct?
Yes.

2)What is the potential at a distance of 5 x 10^-10 m from the nucleus of charge +60e?
I donno if my formula is correct? ids it V=Ed?
No, that formula only works for constant, uniform electric field. What you have here is a Coulomb potential, which you should be able to find in your book.
 

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