# Estimate the capacitance?

jena
Hi,

My question: A homemade capacitor is assembled by placing two 9-in pie pans 5cm apart and connecting them to the opposite terminals of 9-V batter. Estimate the capacitance?

I know that to get the capacitance, you can use C=Q/V or C=keA/d. Do I need to find the area of the two pie pans to be help me solve the problem or do I make a triangle and find the area of that.

Thank You

## Answers and Replies

Staff Emeritus
Since C=Q/V, unless one knows Q, then one will have to use C=keA/d.

I think one can determine the area of a 9-in pie pan. Perhaps assume a circular pie-pan, i.e. 9-in dia.

jena
Hi,

Work

C=keA/d

C=(8085 x 10^-12)(4.104 x 10^-2)/(5 x 10^-2)

C= 7.26 x 10 ^-12

Okay, I see

Thank You

Is C the charge? or Capacitance? Is it C = Q/ deltaV ?

If you used pie pans, could they be filled with something to change the capacitance? like water?

Is the capacitance value increased as the distance gets smaller between the plates?

Staff Emeritus
Brad, C is the Capacitance, Q is the charge.

Putting a material between the conductive plates (some nonconducting (i.e. insulator) dielectric) will affect the capacitance.

C=keA/d indicates that capacitance increases with increasing area and or decreasing distance.

Okay, let's say you have two 5" square sheets of aluminum foil, and put regular paper in between, glue all three items together. The paper is just over 5" sq like 5.1" sq so the foil won't short out. That setup has some capacitance, C.

Then you have two 5" square, well cubed blocks of aluminum, and put the same 5.1" paper between them. The distance is the same (assume no crushing of paper).

Is the capacitance different? What "electrically" would be different?

Staff Emeritus
With more surface area (including the sides), the capacitance of the blocks should be greater than that of the thin foil - given the same square surfaces.

For a given total charge - the surface charge density would be lower with more surface area.