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

Click For Summary

Homework Help Overview

The discussion revolves around calculating various properties of a homemade capacitor, including capacitance, charge, electric field, and work done by the battery. The subject area is primarily focused on electrostatics and capacitor theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate capacitance and charge using provided formulas but struggles with determining the electric field. Participants question the calculations and suggest considering how voltage is distributed between the plates.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the voltage distribution and its effect on the electric field calculation. Some guidance has been offered regarding the use of the equation V = Ed, but no consensus has been reached on the correct approach or final values.

Contextual Notes

The problem involves specific measurements and assumptions about the capacitor's setup, including the distance between the plates and the voltage applied. There may be constraints related to significant figures in the calculations.

mer584
Messages
37
Reaction score
0

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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K