Capacitance of two parallel plate capacitors

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SUMMARY

The capacitance of two parallel plate capacitors, each with square plates of side length 0.1 m and separated by a distance of 10^-3 m in a vacuum, is calculated using the formula C = E0A/L. The area A is determined as 0.01 m², leading to a capacitance value of 8.85 x 10^-11 Farads. The discussion clarifies that the area of the plates is derived from their dimensions, while the separation distance is distinct and does not influence the area calculation.

PREREQUISITES
  • Understanding of capacitance and the formula C = Q/V
  • Familiarity with parallel plate capacitor equations, specifically C = E0A/L
  • Knowledge of the physical constants, particularly E0 (permittivity of free space)
  • Basic geometry to calculate the area of square plates
NEXT STEPS
  • Research the concept of permittivity of free space (E0) and its significance in capacitance calculations
  • Learn about the effects of plate area and separation distance on capacitance
  • Explore different capacitor configurations and their capacitance formulas
  • Investigate practical applications of capacitors in electronic circuits
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Students studying electrical engineering, physics enthusiasts, and anyone looking to understand the principles of capacitance in parallel plate capacitors.

Violagirl
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Homework Statement



Two square metal plates with sides of length 0.1 m are separated in vacuum by 10-3 m. Find their capacitance.

Homework Equations



C = Q/V

For a parallel plate capacitor:

C = E0A/L


The Attempt at a Solution



If A is supposed to be the length of the metal plates with 0.1 m, since A has units of m2, I took A = (.1m)2 = .01 m2 and L = 10-3 m.

From there and plugging everything in, I get:

C = (8.85 x 10-12 C2/N m2 * .01m2/10-3 m

C = 8.85 x 10-11 Farads

If someone could confirm whether this is correct or not, I would appreciate it.
 
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Violagirl said:
If A is supposed to be the length of the metal plates with 0.1 m, since A has units of m2, I took A = (.1m)2 = .01 m2 and L = 10-3 m.
A is the area of each plate. (The area of one face of a plate.)
 
Right about A. What I meant was, of the two numerical values that I'm provided, which would be my A? Or how do I determine my A? I know then that my separation distance corresponds to L. So do I need to determine A by my length values in this situation?
 
A is the area of the square plate. It's 0.1m per side, so A = (0.1m)² as you showed.
 
Violagirl said:

Homework Statement



Two square metal plates with sides of length 0.1 m
This is the dimension of plate.
are separated in vacuum by 10-3 m. Find their capacitance.

This is the distance between the plate
Do not get confused by them. Dimensions of the plate has nothing to do with distance between the plates.

(Also the title is wrong there's only one parallel plate capacitor here, two plates make a capacitor.)
 
Enigman said:
This is the dimension of plate.


This is the distance between the plate
Do not get confused by them. Dimensions of the plate has nothing to do with distance between the plates.

(Also the title is wrong there's only one parallel plate capacitor here, two plates make a capacitor.)

Oh shoot! I did not realize that. Do I need to redo this problem then and look at other equations to determine capacitance then?
 
Er...no everything is correct; just show how you found area by the dimensions given.
 
Oh got it! Will do, thanks so much for looking at this!
 

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