How Do You Calculate Capacitance and Electric Potential?

[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 separated 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

 

[quantumdude]

Originally posted by Dx
1) a .2F cap is desired. What area must the pplates have if they are to be separated 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[/color], which you should be able to find in your book.

 

[M98Ranger]

Hello,

For the first question, your formula and calculation are correct. The area of the plates should be approximately 5x10^7 m^2 to achieve a capacitance of 0.2F with a 2.2mm air gap.

For the second question, you are correct that the formula for potential is V=Ed. In this case, E represents the electric field strength, which can be calculated using the formula E=kQ/r^2, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the charge of the nucleus (+60e = 60 times the charge of an electron), and r is the distance from the nucleus (5x10^-10 m). So the potential at this distance would be V= (9x10^9 Nm^2/C^2)(60e)(5x10^-10 m)/ (5x10^-10 m)^2 = 108 V.

I hope this helps! Let me know if you have any other questions.