How Do You Calculate the Capacitance of a Homemade Capacitor?

[csimon1]

Homework Statement

A science-fair radio uses a homemade capacitor made of two 45 45 sheets of aluminum foil separated by a 0.30-mm-thick sheet of paper. Assume dielectric constant of paper is 3.0.

What is its capacitance?

Homework Equations

C=k e0 A / d

The Attempt at a Solution

I'm not sure if this is the wrong formula or if I am supposed to double .45m since there are two plates. Anyways, I came up with this...

(3.0) * (8.85*e-12) * (.45) / .0003

But the answer I came up with is wrong. Can anyone steer me in the right direction?

 

[willem2]

you need the area of the plate in the equation, not the length of the side. 0.45m is the length of the side of the pieces of aluminium foil

 

[dacruick]

boom, what willem said. and next time include units on the 45, because that number means nothing to me

 

[dacruick]

and next time, if you have the answer in front of you and its a simple calculation such as this. divide your answer by the right answer, or vice versa, and you can figure out the factor that you are off by. and that can be very helpful

 

[csimon1]

oh I am sorry, i wrote what i did down wrong. I did do by area -- 45cmx45cm=2025 and that is 20.25 meters. So i multiplied (3.0) * (8.85*e-12) * (20.25m) / .0003 and the answer I got was 1.79*e-6. I know I am messing up a simple calculation somewhere!

 

[Adjuster]

oh I am sorry, i wrote what i did down wrong. I did do by area -- 45cmx45cm=2025 and that is 20.25 meters. So i multiplied (3.0) * (8.85*e-12) * (20.25m) / .0003 and the answer I got was 1.79*e-6. I know I am messing up a simple calculation somewhere!

UNITS! If you are working in SI, then your area needs to be in m²:

45cm x 45cm is equivalent to 0.45m x 0.45m = 0.2025m²

Now try with that.