How Do You Calculate Capacitance and Electric Field for a Homemade Capacitor?

• mer584
In summary, the conversation discussed the assembly of a homemade capacitor using two 9-inch pie pans placed 5cm apart and connected to a 9V battery. The estimated values for capacitance and charge were 7 x 10^-12 and 7 x 10^-11 respectively. The electric field halfway between the plates was calculated to be approximately 200 V/m. The work done by the battery to charge the plates was not explicitly calculated, but was estimated to be close to 200 V/m.
mer584

Homework Statement

A homemade capacitor is assembled by placing two 9inch pie pants 5cm apart and connecting them to the opposite terminals of a 9V battery. Estimate A) the capacitance, B) the charge on each plate c) the electric field halfway between the plates d) the work done by the battery to charge the plates.

Homework Equations

C=Eo(A/d) V=Ed Q=VC W= Q(V/2) PE= 1/2(QV)= 1/2(CV^2)= 1/2(Q^2/C)

The Attempt at a Solution

I was able to use the first formula I listed to solve for part A finding C= 7 x 10^ -12 and the third formula to solve for part Q= 7x 10^-11 but I can't seem to get part c. I tried using V=Ed but it doesn't give me the answer I'm looking for which should be 200 V/m.

What answer are you getting? (What if you round it off?)

I was getting 360 but then I realized that you probably want to divide the 9/2 since the voltage will be split evenly between each plate so then you end up getting 180..which may be close enough to 200??

mer584 said:
I was getting 360 but then I realized that you probably want to divide the 9/2 since the voltage will be split evenly between each plate
Use the equation V = Ed to solve for E.
so then you end up getting 180..which may be close enough to 200??
That's what I'd say. 180 rounded to one significant figure is 200.

1. What is an electric field?

An electric field is a region around a charged particle or object where other charged particles will experience a force. It is a vector quantity, meaning it has both magnitude and direction.

2. How is an electric field created?

An electric field is created by a charged particle or object. The strength of the electric field is determined by the amount of charge and the distance from the charged particle or object.

3. What is the unit of electric field?

The unit of electric field is newtons per coulomb (N/C) in the SI system of units. In the CGS system, the unit is dynes per statcoulomb (dyn/cm^2).

4. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in farads (F) and is determined by the geometry of the system and the material properties of the conductors and insulators.

5. How does capacitance affect electric fields?

Capacitance affects electric fields by storing energy in the form of an electric field between two conductors. The larger the capacitance, the more energy can be stored in the electric field. Additionally, the presence of a capacitor can also alter the electric field in its vicinity.

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